CHAP CHAPTE TER R4
MOTIO OTION N IN MOR MORE E THAN THAN ONE ONE DIMENSION
ActivPhysics can help with these problems: All Activities in Section 3, Projectile Motion
Section Section 4-1: 4-1:
Velocity Velocity and Acceler Acceleration ation
Problem 1. A skater skater is gliding along the the ice at 2.4 m/s, when she undergoes an acceleration of magnitude 1.1 m/s2 for 3.0 s. At the end of that time she is moving at 5.7 m/s. What must be the angle between the acceleration vector and the initial velocity vector?
Solution For constant acceleration, Equation 4-3 shows that the vectors v 0 , a∆t and a nd v form a triangle as shown. The law of cosines gives v 2 = v 02 + (a ( a∆t)2 2v0 a∆t cos(180 θ0 ). When the given magnitudes are substituted, one can solve for θ 0 : (5. (5.7 m/s)2 = 2 (2. (2.4 m/s)2 + (1. (1.1 m/s )2 (3. (3.0 s)2 + 2(2. 2(2.4 m/s) 2 (1. (1.1 m/s )(3. )(3.0 s)cos θ0 , or cos θ0 = 1.00 (exactly), and θ0 = 0 . Since v0 and a are colinear, the change in speed is maximal. maximal. ◦
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Problem 3. An object is moving in the x direction at 1.3 m/s when it is subjected to an acceleration given by 2 a = 0.52ˆ m/s . What is its velocity vector after 4.4 s of acceleration? acceleration?
Solution From Equation 4-3, v = v0 + a∆t = (1. (1.30 m/s)ˆı + 2 (0. (0.52ˆ m/s )(4. )(4.4s) = (1. (1 .30ˆı + 2.29ˆ) m/s. m/s.
Problem 4. An airliner airliner is flying at a velocity velocity of 260ˆı m/s, when a wind gust gives it an acceleration of 0 .38ˆı + 0.72ˆ m/s 2 for a period of 24 s. (a) What What is its velocity at the end of that time? (b) By what angle has it been deflected from its original course?
Solution (a) Equation 4-3 gives v = 260ˆı m/s + (0. (0.38ˆı + 2 0.72ˆ)(m/ )(m/s )(24 s) = (269ˆı + 17 17..3ˆ) m/s. m/s. (b) Since v0 is along the x the x-axis -axis,, the angular deflection deflection is just tan 1 (vy /vx) = tan 1 (17. (17.3/269) = 3. 3.67 . −
Section Section 4-2:
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Constant Constant Accelera Acceleration tion
Problem 5. The position of an object as a function of time is given by r = (3. (3 .2t + 1. 1.8t2 )ˆı + (1. (1.7t 2.4t2 )ˆ m, where t is is the time in seconds. What are the magnitude and direction of the acceleration?
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Problem 1 Solution.
Solution Problem 2. In the preceding preceding problem, problem, what would have been the magnitude of the skater’s final velocity if the acceleration had been perpendicular to her initial velocity?
Solution When v 0 and a ∆t are perpendicular, Equation 4-3 and the Phythagorean theorem imply v = v02 + (a ( a∆t)2 =
(2. (2.4
m/s)2
2
+ (1. (1.1 m/s )(3. )(3.0
s)2
= 4. 4 .08 m/s. m/s.
One can always find the acceleration by taking the second derivative of the position, a (t) = d 2 r(t)/dt2 . However, collecting terms with the same power of t, one can write the position in meters as r(t) = (3. (3.2ˆı + 1.7ˆ)t + (1. (1 .8ˆı − 2.4ˆ)t2 . Comparison with Equation 4-4 shows that this represents motion with constant acceleration equal to twice the coefficient of the t 2 term, or a = (3. (3.6ˆı − 4.8ˆ) m/s2 . The magnitude a are and direction of a (3. (3.6)2 + ( 4.8)2 m/s2 = 4.49 m/s2 , and tan 1 ( 4.8/3.6) = 307 (CCW from the x the x-axis, -axis, in the fourth quadrant) or 53 53..1 (CW from the x the x-axis). -axis). −
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Problem 6. An airplane heads northeastward down a runway, runway, accelerating from rest at the rate of 2.1 m/s 2. Express the plane’s velocity and position at t = t = 30 s in unit vector notation, notation, using a coordinate coordinate system y with x with x-axis -axis eastward and -axis -axis northward, and with origin at the start of the plane’s takeoff roll.
Solution Since the acceleration is constant and the airplane starts from rest ( v0 = 0) at the origin ( r0 = 0), 0), Equations 4-3 and 4-4 give v(t) = at and r (t) = 21 at2 . Both vectors are in the NE direction, parallel to a = (2. (2.1 m/s2 )(ˆı cos45 +ˆ sin45 ) = (1. (1.48 m/s2 ) (ˆı + ˆ). Thus, v(30 s) = (1. (1 .48 m/s2 )(30 s)(ˆı + ˆ) = (44. (44.5 m/s)(ˆı + ˆ), and r(30 s) = 21 (1. (1.48 m/s)(30 s)2 (ˆı + ˆ) = (668 m) (ˆı + ˆ). (Note: (ˆı + ˆ)/ 2 is a unit vector in the NE direction.) ◦
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Problem
displacement during the rocket firing, not the pathlength of its trajectory.)
Problem 8. An object is moving initial initially ly in the x direction at 4.5 m/s, when an acceleration is applied in the y direction for a period of 18 s. If it moves equal distances in the x and y directions directions during this time, what is the magnitude magnitude of its accelerati acceleration? on?
Solution The x The x component of the displacement is due only to the initial velocity, ∆x ∆ x = v = v0x ∆t. The component is t. The y component just due to the acceleration, ∆ y = 21 a∆t2 = ∆x. ∆ x. Thus, Thus, 2 2(4.5 m/s)/ m/s)/18 s = 0. 0 .5 m/s . a = 2v0x /∆t = 2(4.
Problem 9. A hockey puck is moving at 14. 14 .5 m/s when a stick imparts a constant acceleration of 78. 78 .2 m/s2 at a 90 90..0 angle to the original direction of motion. If the acceleration lasts 0.120 s, what is the magnitude of the puck’s displacement during this time? ◦
7. An asteroid is heading heading toward Earth Earth at a steady 21 km/s. To save save their planet, astronauts astronauts strap a giant rocket to the asteroid, giving it an acceleration of 0.035 km/s 2 at right angles to its original motion. If the rocket firing lasts 250 s, (a) by what angle angle does the direction direction of the asteroid’s motion change? (b) How far does it move during the firing?
Solution
Solution Take the x the x-axis -axis in the direction of the initial velocity, v0 = 14 14..5 m/s, and the y -axis -axis in the direction of the acceleration, a = 78 78..2ˆ m/s 2 . The displacement during the 0.120 s interval of constant acceleration is (Equation 4-4) ∆r = r − r0 = v0 t + 21 at2 = (14. (14.5 m/s)(0. m/s)(0.120 s)ˆı + 21 (78. (78.2 m/s2 )(0. )(0.120 s)2ˆ = (1. (1.74ˆı + 0.563ˆ) m. This has magnitude (1. (1.74)2 + (0. (0.563)2 m = 1.83 m (and makes an angle 1 of tan (0. (0.563 563//1.74) = 17. 17.9 with the direction of the initial velocity).
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Problem 10. Repeat the preceding preceding problem, except except that now the acceleration makes a 65.0 angle with the original direction of motion. ◦
figure
4-2 a Problem Problem 7 Solution.
Solution (a) The change in velocity, ∆ v = a∆t, is t, is at right angles to the initial velocity, v 0 , so they form the legs of a right triangle with hypotenuse v (see Fig. 4-2 a ). ). The angle between v and v0 is θ is θ = tan 1 ( ∆v v0 ) = tan 1 (0. (0.035 km/s)(250 s)/ s)/(21 km/s) = 22. 22 .6 . (b) For constant acceleration, the displacement is (see Equation 4-4) ∆ r = r − r0 = v0 t + 21 at2 . Again, the two vectors v0 t and and 21 at2 form the legs of a right triangle, whose hypotenuse, ∆ r, has magnitude −
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(21 km/s)2 (250 s)2 + ( 12 )2 (0. (0.035 km/s2 )2 (250 s)4 = 5.36 103 km km.. (Note: this is the asteroid’s
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Now the accelerati acceleration on is (78. (78.2 m/s2 )ˆı cos65. cos65.0 + ˆ sin65. r sin65.0 ) and ∆ = (14. (14.5 m/s)(0. m/s)(0.120 s)ˆı + 2 1 (33. (33.0ˆı + 70 70..9ˆ)(m/s ) (0. (0.120 s)2 = (1. (1 .98ˆı + 0.510ˆ) m. 2 This has magnitude 2.04 m (and makes an angle of 14 14..5 with the initial velocity). ◦
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Problem 11. A particle particle leaves leaves the origin origin with initial initial velocity velocity v0 = 11ˆı + 14ˆ m/s. m/s . It undergoes a constant acceleration given by a = − 1.2ˆı + 0.26ˆ m/s 2 . (a) When does the particle cross the y -axis? -axis?
CHAPTER 4
(b) What is its y -coordinate at the time? (c) How fast is it moving, and in what direction, at that time?
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acceleration is assumed constant, the electron trajectory is parabolic in the deflecting region. y
Solution
Deflecting region 4.2 cm
(a) Since the particle leaves from the origin ( r0 = 0), its position is r(t) = v 0 t + 21 at2 . It crosses the y -axis when x(t) = v 0x t + 21 ax t2 = 0, or t = 2v0x /ax = 2(11 m/s)/( 1.2 m/s2 ) = 18.3 s. (b) y(t) = v 0y t + 1 a t2 = [14 m/s + 21 (0.26 m/s2 )(18.3 s)](18.3 s) = 2 y 300 m. (c) v(t) = v0 + at = (11ˆı + 14ˆ) m/s + ( 1.2ˆı+ 0.26ˆ)(18.3)m/s = ( 11ˆı + 18.8ˆ) m/s. Then v(t) = ( 11)2 + (18.8)2 m/s = 21.8 m/s and θx = tan 1 (18.8/( 11)) = 120 .
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figure
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4-27 Problem 13 Solution.
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Section 4-3:
Problem 12. A particle starts from the origin with initial velocity v0 = v0ˆı and constant acceleration a = aˆ. Show that the particle’s distance from the origin and its direction relative to the x-axis are given by d = t
Electron gun
v02 + 41 a2 t2 and θ = tan
1
−
(at/2v0 ).
Projectile Motion
Problem 14. You toss an apple horizontally at 8.7 m/s from a height of 2.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground?
Solution
Solution
From Equation 4-4, the particle’s displacement from its original position is ∆ r = r − r0 = v0 t + 21 at2 =
The time of flight for either projectile can be determined from the vertical component of the motion, which is the same for both, since v 0y = 0. Thus, Equation 4-8 gives t = 2(y y0 )/g =
(v0ˆı + 21 at ˆ)t. This has magnitude t
v02 + ( 12 at)2 and
direction (CCW from the x-axis) θ = tan
1
−
(at/2v0 ).
Problem 13. Figure 4-27 shows a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientific instruments. Electrons are accelerated by the electron gun, then move down the center of the tube at 2.0 109 cm/s. In the 4.2-cm-long deflecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration “steers” them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is needed to deflect the electrons through 15 , as shown in the figure? (b) What is the shape of an electron’s path in the deflecting region?
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2(2.6 m)/(9.8 m/s2 ) = 0.728 s.
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Problem 15. A carpenter tosses a shingle off a 8.8-m-high roof, giving it an initially horizontal velocity of 11 m/s . (a) How long does it take to reach the ground? (b) How far does it move horizontally in this time? y v
0
(11ms)î
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y0 8.8 m
Solution (a) With x-y axes as drawn on Fig. 4-27, the electrons emerge from the deflecting region with velocity v = v0ˆı + atˆ , after a time t = x/v0 , where x = 4.2 cm and v 0 = 2 109 cm/s. The angle of deflection (direction of v) is 15 , so tan15 = v y /vx = at/v0 = ax/v02 . Thus, a = v 02 tan 15 /x = 2.55 1017 cm/s2 (when values are substituted). (b) Since the
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x
Problem 15 Solution.
x
CHAPTER 4
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Solution (a) The shingle reaches the ground when y(t) = 0 = y0 21 gt 2 , or
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t =
2y0 = g
2(8.8 m) 2
(9.8 m/s )
= 1.34 s.
(b) The horizontal displacement is x = v 0 t = (11 m/s)(1.34 s) = 14.7 m.
Problem 16. An arrow fired horizontally at 41 m/s travels 23 m horizontally before it hits the ground. From what height was it fired?
Solution From x x0 = v 0x t, one finds the time of flight t = 23 m/(41 m/s) = 0.561 s, and from y 0 y = 21 gt2 (recall that v 0y = 0) one finds the height y 0 y = 1 (9.8 m/s2 )(0.561 s)2 = 1.54 m. 2
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Problem 17. A kid fires water horizontally from a squirt gun held 1.6 m above the ground. It hits another kid 2.1 m away square in the back, at a point 0.93 m above the ground (see Fig. 4-28). What was the initial speed of the water?
land on another roof 1.9 m lower. If the gap between the buildings is wide, how fast must he run?
Solution The horizontal and vertical distances covered by the stuntman are x x0 = v 0 t and y 0 y = 21 gt 2 (since v 0x = v 0 , and v 0y = 0). Eliminating t , one finds v 0 = (x x0 ) g/2(y0 y) =
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(4.5 m) (9.8 m/s2 )/2(1.9 m) = 7.23 m/s. (Note that Equation 4-9 with θ 0 = 0 and y 0 = 0 provides an equivalent solution.)
Problem 19. Ink droplets in an ink-jet printer are ejected horizontally at 12 m/s, and travel a horizontal distance of 1.0 mm to the paper. How far do they fall in this interval?
Solution From x x0 = v 0x t, the time of flight can be found. Substitution into y 0 y = 21 gt2 (recall that v0y = 0) yields y 0 y = 21 g(x x0 )2 /v02 = 1 (9.8 m/s2 )(10 3 m)2 /(12 m/s)2 = 3.40 10 8 m = 2 34 nm for the distance fallen, practically negligible. Note that this analysis is equivalent to using Equation 4-9 with θ 0 = 0.
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Problem 20. Protons in a particle accelerator drop 1 .2 µm over the 1.7-km length of the accelerator. What is their approximate average speed?
1.6 m 0.93 m
2.1 m figure
4-28 Problem 17.
Solution The horizontal and vertical distances in projectile motion (range and drop) are related by the trajectory equation (Equation 4-9). With θ 0 = 0, y = gx 2 /2v02 , or v 0 = x ( g)/2y =
3
6
Solution
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Problem 18. In a chase scene, a movie stuntman is supposed to run right off the flat roof of one city building and
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3
2(1.6 0.93) m/(9.8 m/s2 ) = 0.370 s. Its initial speed, v 0 = v 0x , can be found from Equation 4-7, v0 = (x x0 )/t = 2.1 m/0.370 s = 5.68 m/s.
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Since the water was fired horizontally ( v0y = 0), the time it takes to fall from y 0 = 1.6 m to y = 0.93m is given by Equation 4-8, t = 2(y0 y)/g =
− − (1.7×10 m) (−9.8 m/s )/2(−1.2×10 m) = 3.44×10 m/s. Since v y = −gt = −g(x − x )/v = −4.85×10 m/s is negligible compared to this, v 0
0
0
is
hardly different from the average speed.
Problem 21. You’re standing on the ground 3.0 m from the wall of a building, and you want to throw a package from your 1.5-4.2m shoulder level to someone in a second-floor window above the ground. At what speed and angle should you throw it so it just barely reaches the window?
CHAPTER 4
Solution
impact, was the pedestrian in the crosswalk?
We suppose that “just barely” means that the maximum height of the package equals the height of the window sill. When the package reaches the sill (in the coordinate system shown), v y2 = 0 = v 02y 2gy,
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so v0y = 2(9.8 m/s2 )(2.7 m) = 7.27 m/s. Since vy = 0 = v 0y gt, the time of flight is t = v 0y /g. Therefore v 0x = x/t = (3.0 m)(9.8 m/s2 )/(7.27 m/s) = 4.04 m/s. From these components, we find:
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v0 = θ0 =
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v02x + v02y = 8.32 m/s, and tan −1 (v0y /v0x ) = 60.9◦ .
Solution What is an issue here is the horizontal range of a piece of signal light lens in projectile motion, starting from a height of y 0 y = 0.63 m off the ground, with an initial horizontal velocity of v 0 = v 0x = (40 m/3.6 s), and v 0y = 0. Eliminating the time of flight from Equations 4-7 and 4-8 (see the solution to Problem 19), one obtains x x0 = v0 2(y0 y)/g =
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(11.1 m/s) 2(0.63 m)/(9.8 m/s2 ) = 3.98 m. If the pieces of lens did not bounce very far from the point where they hit the ground, this places the point of impact of the accident just 4 .00 m 3.98 m = 2 cm from the center of the crosswalk. (Forensic physics is crucial to the prosecution’s case.)
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Problem 24. Repeat Problem 15 for the case when the shingle is thrown with a speed of 11 m/s at 14 below the horizontal. ◦
Solution (a) If the initial velocity of the shingle has a downward component, v 0y = v 0 sin θ0 = (11 m/s) sin( 14 ) = 2.66 m/s, the time t to reach the ground is now given by the positive solution (t > 0) of the equation y (t) = 0 = y 0 + v0y t 21 gt 2. From the quadratic formula, this is t = (v0y +
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Problem 21 Solution.
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v02y + 2gy 0 )/g = [ 2.66 m/s +
22. Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed v 0 from a height h .
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( 2.66 m/s)2 + 2(9.8 m/s2 )(8.8 m)]/(9.8 m/s2 ) = 1.10 s (b) The horizontal range is x x0 = v 0x t = (11 m/s)(cos( 14 ))(1.10 s) = 11.7 m.
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Solution This problem is solved by reasoning identical to that in Problem 15, where it was shown that x = v 0 2h/g. (By referring the student to a similar problem, the wide applicability of the laws of projectile motion is better appreciated.)
Problem 23. A car moving at 40 km/h strikes a pedestrian a glancing blow, breaking both the car’s front signal light lens and the pedestrian’s hip. Pieces of the lens are found 4.0 m down the road from the center of a 1.2-m-wide crosswalk, and a lawsuit hinges on whether or not the pedestrian was in the crosswalk at the time of the accident. Assuming that the lens was initially 63 cm off the ground, and that the lens pieces continued moving horizontally with the car’s speed at the time of the
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Problem 25. In part (b) of the exercise following Example 4-5, what is the vertical component of the velocity with which the cyclist strikes the ground?
Solution One first needs to work out part (a) of the exercise, which is similar to the first part of Example 4-5. At the minimum speed at takeoff, the cyclist covers a horizontal range x = 48 m, and a vertical drop y = 5.9 m (recall that x 0 = y 0 = 0 in Equation 4-9). Then
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v0,min =
2
2
(9.8 m/s )(48 m) 2 2(cos 15 )[(48 m) tan 15 ( 5.9 m)]
= 25.4 m/s.
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The cyclist’s actual initial speed is v 0 = (1 + 50%)(25.4 m/s) = 38.1 m/s. Equation 2-11 gives the vertical component of the velocity (negative downward) for the given drop, v y =
− −
(38.1 m/s)2 sin2 15 14.6 m/s
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−
v02y
− 2gy =
Equation 4-8 when y(t) y0 = 0 = v 0y t 21 gt2 . Thus t = 2v0y /g = 2(6.64 km/s) sin 45 /(9.8 m/s2 ) = 958 s = 16.0 min. (c) At a 20 launch angle,
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v0 =
2
− 2(9.8 m/s )(−5.9 m) =
(4500 km)(0.0098 km/s2 )/ sin40 = 8.28 km/s. ◦
Problem 28. A rescue airplane is flying horizontally at speed v 0 at an altitude h above the ocean, attempting to drop a package of medical supplies to a shipwreck victim in a lifeboat. At what line-of-sight angle α (Fig. 4-29) should the pilot release the package?
Problem 26. Compare the travel times for the projectiles launched at 30 and 60 in Fig. 4-15, both of which have the same starting and ending points. ◦
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Solution The package, when dropped, is a horizontally launched projectile, like that in Problem 15 (the expressions for t and x are the same). The line-of-sight angle is α = tan 1 (h/x), where x = v 0 2h/g is the horizontal distance at drop time. Thus, α = tan 1 gh/2v02 .
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v
0
v0 î
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A
figure
α
4-15 Problem 26.
y0
Solution The time of flight (t > 0) for a projectile trajectory beginning and ending at the same height (y(t) = y 0 ) can be found from Equation 4-8, y(t) y0 = 0 = v 0y t 1 gt 2 or t = 2v0y /g = 2v0 sin θ0 /g. In Fig. 4-15, v 0 = 2 50 m/s, so t 30 = 2(50 m/s) sin30 /(9.8 m/s2 ) = 5.10 s, and t 60 = 3 t 30 = 8.84 s.
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Problem 27. A submarine-launched missile has a range of 4500 km. (a) What launch speed is needed for this range when the launch angle is 45 ? (b) What is the total flight time? (c) What would be the minimum launch speed at a 20 launch angle, used to “depress” the trajectory so as to foil a space-based antimissile defense? ◦
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4-29 Problem 28 Solution.
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Solution (a) Assuming Equation 4-10 applies (i.e., the trajectory begins and ends at the same height, or y(t) = y 0 ), one finds v 0 = xg/ sin2θ =
(4500 km)(0.0098 km/s2 )/ sin90 = 6.64 km/s. (b) The time of flight is the positive solution of ◦
Problem 29. At a circus, a human cannonball is shot from a cannon at 35 km/h at an angle of 40 . If he leaves the cannon 1.0 m off the ground, and lands in a net 2.0 m off the ground, how long is he in the air? ◦
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Solution The time of flight can be calculated from Equation 4-8, where the trajectory begins at y0 = 1.0 m and t = 0, ends at y (t) = 2.0 m, and v0y = v 0 sin θ0 = (35 m/3.6 s)sin40 = 6.25 m/s. The equation is a quadratic, 21 gt2 v0y t + (y y0 ) = 0, ◦
with solutions t = [v0y [6.25 m/s
±
− ± −
(6.25 m/s)2
v02y
−
2g(y
− y )]/g = − 2(9.8 m/s )(2 m − 1 m)]÷ 0
2
(9.8 m/s2 ) = 0.188 s or 1.09 s. The trajectory crosses the height 2.0 m twice, once going up, at the smaller time of flight, and once going down into the net, at the larger time of flight. The latter is the answer to the question asked here.
Problem 30. Preparing to give an injection, a physician ejects a drop of medicine to ensure there’s no air in the syringe. The syringe is pointing upward, at 20 to the vertical, with the tip of the needle 45.0 cm above a tabletop. The drop leaves the needle at 1.30 m/s. (a) What maximum height does it reach? (b) How long is it before the drop hits the tabletop? (c) How far does the drop travel horizontally? ◦
Solution (a) Take the tip of the syringe to be at y 0 = 45.0 cm above the tabletop at y = 0. At its maximum height, just the vertical component of the drop’s velocity is instantaneously zero, so v y2 = 0 = v 02y 2g(ymax y0 ), or y max y0 = v 02y /2g. The vertical component of the initial velocity is v 0y = (1.30 m/s) cos 20 , which gives ymax y0 = 7.61 cm. This is the maximum height above the tip, corresponding to 52.6 cm above the tabletop. (b) The time of flight is the positive solution of the equation y(t) = 0 = y 0 + v0y t 21 gt 2 , or t =
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(v0y + v02y + 2gy 0 )/g. Substituting the numbers from part (a), we obtain t = 0.452 s. (c) The horizontal distance traveled is x x0 = v 0x t = (1.30 m/s)(sin 20 )(0.452 s) = 20.1 cm. ◦
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Problem 31. If you can hit a golf ball 180 m on Earth, how far can you hit it on the moon? (Your answer is an underestimate, because the distance on Earth is restricted by air resistance as well as by a larger g .)
Solution For given v 0 , the horizontal range is inversely proportional to g . With surface gravities from
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Appendix E, we find x moon = (gEarth/gmoon )xEarth = (9.81/1.62)(180 m) = 1090 m.
Problem 32. Prove that a projectile launched on level ground reaches its maximum height midway along its trajectory.
Solution From the trajectory (Equation 4-9), the maximum height occurs when dy/dx = tan θ gx/v02 cos2 θ0 = 0, or x = v02 sin θ0 cos θ/g, which is midway, or half of the horizontal range (Equation 4-10). This result can be derived in other ways; for example, the maximum height is reached at time v 0y /g (when v y = 0), whereas the total time of flight is twice this (the solution of 0 = v 0y t 21 gt2 ).
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Problem 33. A projectile launched at an angle θ 0 to the horizontal reaches a maximum height h . Show that its horizontal range is 4 h/ tan θ0 .
Solution The intermediate expression for the horizontal range (when the initial and final heights are equal) is x = 2v02 sin θ0 cos θ0 /g = 2v0x v0y /g (see the equation before Equation 4-10). The components of the initial velocity are related by v 0y /v0x = tan θ0 . The maximum height, h = y max y0 , can be found from Equation 2-11 (when v y = 0) or v 02y = 2gh. Then x = 2v0y v0x /g = 2v0y (v0y / tan θ0 )/g = 2(2gh)/g tan θ0 = 4h/ tan θ0 . (This result reflects a classical geometrical property of the parabola, namely, that the latus rectum is four times the distance from vertex to focus.)
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Problem 34. You’re 5.0 m from the left-hand wall of the house shown in Fig. 4-30, and you want to throw a ball to a friend 5.0 m from the right-hand wall. (a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle should it be thrown? Assume the throw and catch both occur 1.0 m above the ground.
Solution Since the trajectory is symmetrical (begins and ends at the same height), one can use the result of the previous problem with h = y max y0 = 6 m 1 m = 5 m and horizontal range x = 5 m + 6 m + 5 m = 16 m. Then (b) θ 0 = tan 1 (4h/x) = 51.3 , and (a) v 0 = v 0y / sin θ0 = 2gh/ sin θ0 = 12.7 m/s.
−
√
−
−
◦
28
CHAPTER 4
45
jumped at sea level, where g = 9.81 m/s2 , at the same angle and initial speed as in Mexico City, how far would Beamon have gone? Neglect air resistance in both cases (although its effect is actually more significant than the change in g ).
°
3.0 m 1.0 m
1.0 m
5.0 m
6.0 m
Solution
5.0 m
The length of the jump is inversely proportional to g (just as in Problem 31), so x = (g/g )x = (9.786/9.810)(8.90 m) = 8.88 m. ′
figure
4-30 Problem 34.
′
Problem Problem 35. A circular fountain has jets of water directed from the circumference inward at an angle of 45 . Each jet reaches a maximum height of 2.2 m. (a) If all the jets converge in the center of the circle and at their initial height, what is the radius of the fountain? (b) If one of the jets is aimed at 10 too low, how far short of the center does it fall? ◦
◦
Solution range, r = v 02 /g
(a) The radius is the horizontal (Equation 4-10 with θ = 45 ). The maximum height is h = v 02y /2g = v 02 /4g (Equation 2-11 with v y = 0 and v0y = v 0 cos45 = v 0 / 2). Therefore, r = (4gh)/g = 4h = 4(2.2 m) = 8.8 m. (b) If one jet is directed at 35 with the same initial speed ( v02 = rg), it would fall short by r x, where x is given by Equation 4-10. Therefore, r x = r (v02 /g) sin(2 35 ) = (8.8 m)(1 sin70 ) = 0.531 m. ◦
√
◦
◦
− − −
◦
−
×
◦
37. In 1991 Mike Powell shattered Bob Beamon’s 1968 world long jump record with a leap of 8.95 m (see Fig. 4-31). Studies show that Powell jumps at 22 to the vertical. Treating him as a projectile, at what speed did Powell begin his jump? ◦
Solution The horizontal range formula (Equation 4-10) gives v 0 = xg/ sin2θ0 =
(8.95 m)(9.8 m/s2 )/ sin 2(90 22 ) = 11.2 m/s (Air resistance and body control are important factors in the long jump also.) ◦
−
◦
Problem 38. A motorcyclist driving in a 60-km/h zone hits a stopped car on a level road. The cyclist is thrown from his bike and lands 39 m down the road. Was the cyclist speeding? To answer, find the minimum speed he could have been going just before the accident.
Solution If the motorcyclist was deflected upward from the road at an angle of 45 , the horizontal range formula (Equation 4-10) implies a minimum initial speed of ◦
Problem 35 Solution.
Problem 36. When the Olympics were held in Mexico City in 1968, many sports fans feared that the high altitude would result in poor performances due to reduced oxygen. To their surprise, new records were set in track and field events, probably as a result of lowered air resistance and a decrease in g to 9.786 m/s2 , both ultimately associated with the high altitude. In particular, Robert Beamon set a new world record of 8.90 m in the long jump. Photographs suggest that Beamon started his jump at a 25 angle to the horizontal. If he had ◦
√ v = xg =
(39 m)(9.8 m/s2 ) = 19.5 m/s = 70.4 km/h. In fact, some speed would be lost during impact with the car, so the cyclist probably was speeding. 0
Problem 39. Show that, for a given initial speed, the horizontal range of a projectile is the same for launch angles 45 + α and 45 α, where α is between 0 and 45 . ◦
◦
◦
◦
−
Solution The trigonometric identity in Appendix A for the sine of the sum of two angles shows that sin 2(45 α) = sin(90 2α) = sin90 cos2α cos90 sin2α = cos 2α, ◦
◦
±
◦
±
◦
±
CHAPTER 4
so the horizontal range formula (Equation 4-10) gives the same range for either launch angle, at the same initial speed.
Problem 40. One model of the Scud missile used in the 1991 Persian Gulf war has a range of 630 km. (a) What is its launch speed, given a 40 launch angle? (b) What is the missile flight time? ◦
Solution (a) From Equation 4-10, v 0 =
xg/ sin2θ0 =
(630 km)(.0098 km/s2 )/ sin80 = 2.50 km/s. (b) From Equation 4-7, the time of flight is t = x/v0 cos 40 = (630 km)/(2.50 km/s) cos 40 = 328 s = 5.47 min . (We assumed the same initial and final trajectory elevations and neglected air resistance.) ◦
◦
◦
Problem 41. A basketball player is 15 ft horizontally from the center of the basket, which is 10 ft off the ground. At what angle should the player aim the ball if it is thrown from a height of 8.2 ft with a speed of 26 ft/s?
Solution With origin at the point from which the ball is thrown, the equation of the trajectory (Equation 4-9), evaluated at the basket, becomes (32 ft/s 2 )(15 ft)2 y = (10 8.2)ft = (15 ft) tan θ0 , 2(26 ft/s)2 cos2 θ0 or 1.8 = 15 tan θ0 5.33/ cos2 θ0 . Using the trigonometric identity 1 + tan 2 θ0 = 1/ cos2 θ0 , we can convert this equation into a quadratic in tan θ0 : 7.13 15 tan θ0 + 5.33tan2 θ0 = 0, so
−
−
−
−
θ0 = tan
1
−
◦
±
152 4(5.33)(7.13) 2(5.33)
15
= 31.2 or 65.7
−
◦
Like the horizontal range formula (see Fig. 4-13), for given v 0 there are two launch angles whose trajectories
Problem 41 Solution.
29
pass through the basket, although in this case they are not symmetrically placed about 45 . ◦
30
CHAPTER 4
Section 4-4: Circular Motion
α in radians before dividing.) Most calculus texts contain a proof that (sin x/x) 1 for x 0.
→
Problem
→
42. How fast would a car have to round a turn 75 m in radius in order for its acceleration to be numerically equal to that of gravity?
Solution For circular motion with constant speed, a r = v 2 /r or
√
v = ar r = (9.8 m/s2 )(75 m) = 27.1 m/s = 97.6 km/h = 60.7 mi/h.
Problem 44 Solution.
Problem 43. Estimate the acceleration of the moon, which completes a nearly circular orbit of 385,000 km radius in 27 days.
Problem
Solution The centripetal acceleration is given in terms of the period for uniform circular motion by Equation 4-12 in Example 4-8. In the case of the moon, a = 4π2 r/T 2 = 4π 2 (3.85 108 m)/(27.3 86, 400 s)2 = 2.73 10 3 m/s2 , where we used more accurate data from Appendix E. (Note: “centripetal” is a purely kinematic adjective descriptive of circular motion. In this case, the origin of the moon’s centripetal acceleration is the gravitational attraction of the Earth.)
×
×
−
×
Solution Consider a circular orbit around the moon with radius slightly larger than the lunar radius, r = 1.74 106 m, and centripetal acceleration approximately equal to the lunar surface gravity, a c = 1.62 m/s2 (see Appendix E). The orbital period is related to r and a c by Equation 4-12 as in Example 4-8, T = 2π r/ac , and radio communications with Earth were blocked for half of this period, or 21 T =
×
Problem 44. An object is in uniform circular motion. Make a graph of its average acceleration, measured in units of v 2 /r, versus angular separation for points on the circular path spaced 40 , 30 , 20 , and 10 apart. Your graph should show the average acceleration approaching the instantaneous value v2 /r as the angular separation decreases. ◦
◦
◦
◦
Solution In uniform circular motion, the average acceleration is a multiple of the base of an isosceles triangle with apex angle equal to the angular separation of the positions at the interval’s endpoints (see Fig. 4-21). Using some trigonometry, we find ∆v = 2v2 (1 cos∆θ) = 2v sin( 12 ∆θ). Since, for constant speed, v = (arc length)/(time) = r∆θ/∆t,
45. When Apollo astronauts landed on the moon, they left one astronaut behind in a circular orbit around the moon. For the half of the orbit spent over the far side of the moon, that individual was completely cut off from communication with the rest of humanity. How long did this lonely state last? Assume a sufficiently low orbit that you can use the moon’s surface gravitational acceleration (see Appendix E) for the spacecraft.
| |
−
1 2
∆θ) v |aav | = |∆∆v| = 2v( sin( = ∆θ/v) t
r
2
r
sin( 12 ∆θ) ( 12 ∆θ)
The graph of aav , in units of v 2 /r, versus ∆θ is the same as the graph of f (α) = sin( 12 α)/( 12 α) versus α. For α = 40 , 30 , 20 , and 10 , f (α) = 0.980, 0.989, 0.995, and 0.999 respectively. (Don’t forget to express
| |
◦
◦
◦
◦
π (1.74 106 m)/(1.62 m/s2 ) = 3.26 103 s or about 54.3 min.
×
×
Problem 46. A 12-in-diameter circular saw blade rotates at 3500 revolutions per minute. What is the acceleration of one of the saw teeth? Compare with the acceleration of gravity.
Solution If the saw blade rotates at 3500 rpm, a point on its circumference has a linear speed of v = (3500)π(12 in)/60 s = 183 ft/s. The (radial) acceleration is a r = v 2 /r = (183 ft/s)2 /(6 ft/12) = 6.72 104 ft/s2 2 103 g, where g = 32.2 ft/s2 . Alternatively, one could use Equation 4-12 with period equal to (3500) 1 min .
×
≃ ×
−
CHAPTER 4
31
Problem 47. A jet is diving vertically downward at 1200 km/h (see Fig. 4-32). If the pilot can withstand a maximum acceleration of 5 g (i.e., 5 times Earth’s gravitational acceleration) before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive? Assume the speed remains constant.
r = 4.30 cm °
55
Solution The height at the start of the 90 -turn must be greater than the radius of the turn, in order to avoid hitting the ground. The radius of the turn must be great enough that the centripetal acceleration not exceed 5g , i.e., a c = v 2 /r 5g or r v2 /5 g = (1200 m/3.6 s)2 = 5(9.8 m/s2 ) = 2.27 km. ◦
≤
≥
figure
4-33 Problem 48.
Problem 49. How long would a day last if Earth were rotating so fast that the acceleration of an object on the equator were equal to g ?
Solution
v
The Earth’s equitorial radius is about 6378 km. If the centripetal acceleration at the equator were ac = 9.8 m/s2 , the Earth’s period of rotation would have to be T = 2π RE /ac =
2π (6378 km)/(0.0098 km/s2 ) = 5.07 103 s = 1 h 24.5 min. (See Equation 4-12.)
r
×
Problem
figure
4-32 Problem 47 Solution.
Solution
Problem 48. Electrons in a TV tube are deflected through a 55 angle, as shown in Fig. 4-33. During the deflection they move at constant speed in a circular path of radius 4.30 cm. If they experience an acceleration of 3.35 1017 m/s2 , how long does the deflection take? ◦
×
Solution
50. A runner rounds the semicircular end of a track whose curvature radius is 16 m. The runner moves at constant speed, with an acceleration of 0.94 m/s2 . How long does it take to complete the turn?
√ ar r = m/s )(0.043 m) = 1.20×10
Since the runner has a constant speed along a circular arc, the acceleration must be purely a centripetal acceleration, namely a c = 0.94 m/s2 = v 2 /r, and the
speed is v = (0.94 m/s2 )(16 m) = 3.88 m/s. At this speed, it takes time t = πr/v = π(16 m)/(3.88 m/s) = 13.0 s to complete a semicircle. (Of course, Equation 4-12 would give the same result, 21 T = π r/ac .)
The speed of the electrons is v =
Section 4-5:
Problem
2 8 (3.35 1017 m/s. The length of a circular arc of 55 and radius 0.043 m is 2π(0.043 m)(55/360) = 0.0413 m. Therefore, the time for the deflection is t = 0.0431 m/1.2 108 m/s = 0.344 ns. (This problem will appear more transparent after studying Section 12-1.)
×
◦
×
Nonuniform Circular Motion
51. A space station 120 m in diameter is set rotating in order to give its occupants “artificial gravity.” Over a period of 5.0 min, small rockets bring the station steadily to its final rotation rate of 1 revolution every 20 s. What are the radial and
32
CHAPTER 4
tangential accelerations of a point on the rim of the station 2.0 min after the rockets start firing?
Solution If the rotation rate increases steadily from 0 to 1 revolution in 20 s, over a 5-minute interval, the rotation rate after 2 minutes is (2 /5)(1 rev/20 s) = 1 rev/50 s, or the (instantaneous) period after 2 minutes is 50 s. Thus, by Equation 4-12, the centripetal (radial) acceleration is a c = 4π 2 r/T 2 = 4π 2 (60 m)/(50 s)2 = 0.947 m/s2 . The tangential speed increases steadily from 0 to 2π(60 m)/20 s = 6π m/s, in 5 minutes, so the tangential acceleration is at = (6π m/s)/5 min = 6.28 10 2 m/s2 . (The solution to this problem may appear more straightforward after the discussion of angular motion in Chapter 12.)
×
−
Problem 52. A plane is heading northward when it begins to turn eastward on a circular path of radius 9.10 km. At the instant it begins to turn, its acceleration vector points 22.0 north of east and has magnitude 2.60 m/s2 . (a) What is the plane’s speed? (b) At what rate is its speed increasing? ◦
Solution (a) The east-component of a is the radial acceleration ar = (2.60 m/s2 )cos22 = v 2 /r. Therefore, the speed ◦
is v = (2.60 m/s2 )cos22 (9.1 km) = 148 m/s. (b) The north-component of a is the tangential acceleration (the rate of increase in speed), so at = (2.60 m/s2 )sin22 = 0.974 m/s2 . ◦
◦
Solution The tangential acceleration (in the direction of motion) is perpendicular to the radial acceleration, so the resultant total acceleration (their vector sum) is at 45 between them at the instant when a t = a r = v 2 /r. The linear speed along the circle depends on a t only (since a c is perpendicular to the velocity), so v = a t t for constant a t (provided the object is “set into motion” with v 0 = 0 at t = 0). Thus, a t = (at t)2 /r or t = r/at . ◦
Problem 54. A car moving at 65 km/h enters a curve that describes a quarter turn of radius 120 m. The driver gently applies the brakes, giving a constant tangential deceleration of magnitude 0 .65 m/s2 . Just before emerging from the turn, what are (a) the magnitude of the car’s acceleration and (b) the angle between the acceleration vector and the direction of motion?
Solution The car’s tangential acceleration is constant, a = 0.65 m/s2 along the direction of motion, so Equation 2-11 can be used to determine the speed at the end of the turn, v 2 = v 02 + 2at s. Here, s = 2πr/4 is the linear distance around the quarter turn. Then the centripetal acceleration is a c = v 2 /r = (v02 + πra t )/r = (65 m/3.6 s)2 /120 m + π( 0.65 m/s2 ) = 0.675 m/s2 . (a) The magnitude of the total acceleration is a2c + a2t = 0.937 m/s2 . (b) The angle of the total acceleration with respect to the tangent to the curve is θ = tan 1 (ac /at ) = 134 . (This angle is in the second quadrant because the tangential acceleration is opposite to the direction of motion.)
−
−
−
◦
Paired Problems Problem 55. An alpine rescue team is using a slingshot to send an emergency medical packet to climbers stranded on a ledge, as shown in Fig. 4-34. What should be the launch speed from the slingshot? Problem 52 Solution.
Solution Problem 53. An object is set into motion on a circular path of radius r by giving it a constant tangential acceleration a t . Derive an expression for the time t when the acceleration vector points at 45 to the direction of motion. ◦
If we take the origin of coordinates at the slingshot and the stranded climbers at x = 390 m and y = 270 m, we can use Equation 4-9 for the trajectory to solve for v 0 : x v0 = cos θ0
g 2(x tan θ0
− y)
CHAPTER 4
=
390 m cos70
◦
9.8 m/s2 = 89.2 m/s. 2(390 m tan70 270 m)
·
◦
−
Problem 56. A cat leaps onto a counter 90 cm off the floor, starting 65 cm from the edge of the counter. It leaps at an initial angle of 79 to the horizontal and lands on the counter 22 cm from the edge. What was its initial speed? ◦
Solution
0.87 m cos79
The solution to the previous problem yields h = 21 x = 90 km. (The approximations implicit in this result are less valid for a rocket than for a stone.)
Problem 59. I can kick a soccer ball 28 m on level ground, giving it an initial velocity at 40 to the horizontal. At the same initial speed and angle to the horizontal, what horizontal distance can I kick the ball on a 15 upward slope? ◦
◦
Solution We need to find the intersection of the trajectory of the ball (Equation 4-9) with a 15 slope through the same origin, y = x tan15 . The appearance of the trajectory equation can be simplified by use of the fact that y = 0 when x = 28 m and θ 0 = 40 . Thus, y = 0 = x tan θ0 (g/2v02x )x2 = (28 m)[tan 40 (g/2v02x )(28 m)], or the coefficient (g/2v02x ) equals tan40 /28 m. The trajectory equation simplifies to y = x tan40 x2 (tan 40 /28 m) = x(1 x/28 m)tan40 . The intersection of this with the slope occurs when y also equals x tan15 , or x tan15 = x(1 x/28 m)tan40 . The x-coordinates of the two points of intersection are x = 0 (the origin) and x = (28 m)(1 tan15 / tan40 ) = 19.1 m (the horizontal distance queried in this problem). ◦
2
9.8 m/s 2(0.87 m tan79
·
◦
− 0.9 m)
= 5.34 m/s.
Problem 57. If you can throw a stone straight up to a height of 16 m, how far could you throw it horizontally over level ground? Assume the same throwing speed and optimum launch angle.
Solution
◦
◦
−
−√
◦
−
−
◦
◦
−
◦
−
◦
◦
◦
To throw an object vertically to a maximum height of h = 16 m = y max y0 requires an initial speed of v 0 = 2g(ymax y0 ) = 2gh. With this value of v 0 and the optimum launch angle θ 0 = 45 , Equation 4-10 gives a maximum horizontal range on level ground of x = v 02 /g = 2h = 32 m. (The maximum horizontal range on level ground is twice the maximum height for vertical motion with the same initial speed. This result holds in the approximation of constant g and no air resistance.)
Solution
◦
If the cat jumps from the origin, it lands at x = 65 cm + 22 cm = 0 .87 m and y = 0.9 m. Equation 4-9 solved for v 0 gives v0 =
33
◦
−
◦
−
◦
◦
Problem 59 Solution.
Problem 58. In a conversion from military to peacetime use, a missile with a maximum horizontal range of 180 km is being adapted for studying the upper atmosphere. What is the maximum altitude it can achieve, if launched vertically?
Problem 60. A model rocket has a horizontal range of 280 m on level ground, when given a 45 launch angle. What horizontal distance will the rocket cover when launched at 45 to the horizontal from the top of a hill whose sides slope down at 21 ? ◦
◦
◦
Solution
270 m
70
°
390 m
figure
4-34 Problem 55.
We are given two points on the trajectory (Equation 4-9) of the rocket at the altitude of the hilltop, the origin x = y = 0, and the horizontal range at level x = 280 m and y = 0. As in the previous problem, we seek the x-coordinate of the intersection of the
34
CHAPTER 4
trajectory with the slope y = x tan21 through the origin. Since 0 = (280 m) tan 45 (g/2v02x )(280 m)2 , we can eliminate the constant (g/2v02x ) = 1/280 m from the trajectory equation and find the intersection of it with the slope as before: x tan21 = x x2 (1/280 m), or x = (280 m)(1 + tan 21 ) = 387 m. (Recall that tan45 = 1.)
−
◦
◦
−
◦
−
drops range from 10 below the horizontal to 25 above. If the nozzle is 1.7 m above the ground, how wide is the region (marked w in Fig. 4-35) that gets wet? ◦
−
◦
◦
25
° °
10
◦
w figure
Problem 60 Solution.
4-35 Problem 62.
Solution
Problem 61. A fireworks rocket is 73 m above the ground when it explodes. Immediately after the explosion, one piece is moving at 51 m/s at 23 to the upward vertical direction. A second piece is moving at 38 m/s at 11 below the horizontal direction. At what horizontal distance from the explosion site does each piece land? ◦
◦
Solution In the trajectory equation (Equation 4-9) with origin at the position of the explosion, the coefficients are known for each piece. One can solve this quadratic equation for x, when y = 73 m (ground level), and select the positive root (since the trajectories start at x = 0 and end on the ground in the direction of v 0x , which is chosen positive). For the first piece, tan 23 = 0.424 and g /2v02x = (9.8 m/s2 )/2(51 cos 23 m/s)2 = 2.22 10 3 m 1 , so the quadratic is 73 m = 0.424x (2.22 10 3 m 1 )x2 . This has positive root x = [0.424 + (0.424)2 + 4(73 m)(2.22 10 3 m 1 )] (4.44 10 3 m 1 ) = 300 m. Similarly, for the second piece, tan( 11 ) = 0.194 and g/2v02x = (9.8 m/s2 ) 2(38 cos( 11 ) m/s)2 = 3.52 10 3 m 1 , so x = [ 0.194 + ( 0.194)2 + 4(73 m)(3.52 10 3 m 1 )] (7.04 10 3 m 1 ) = 119 m.
−
We can find where the drops hit the ground from the trajectory equation (Equation 4-9), as in the previous problem. Here, the origin is at the nozzle, and the ground has y = 1.7 m. The positive root of the quadratic 1.7 m = x tan θ0 (g/2v02x )x2 is x = [tan θ0 + tan2 θ0 + 4(1.7 m)(g/2v02x )](v02x /g). When data for the upper and lower extremes of spray (θ0 = 25 or 10 , v0 = 4.6 m/s) are substituted, one finds a spread in x of w = 3.42 m 2.32 m = 1.09 m.
−
◦
−
−
−
◦
−
Problem 63. You toss a chocolate bar to your hiking companion located 8.6 m up a 39 slope, as shown in Fig. 4-36. Determine the initial velocity vector so that the chocolate bar will reach your friend moving horizontally. ◦
◦
◦
× ×
− −
−
−
−
−
×
−
−
−
−
◦
− −
×
÷ −
−
×
× −
−
−
−
÷
−
−
÷
8.6 m
◦
−
×
°
39
figure
4-36 Problem 63.
Problem 62. A hose nozzle sprays water drops in a fan-shaped pattern as suggested in Fig. 4-35. The drops leave the nozzle moving at 4.6 m/s. With the hose aimed as shown, the directions of the emerging
Solution The candy bar moves horizontally only at the apex of its tra jectory (where v x = v 0x and v y = 0). Thus,
CHAPTER 4
− y = (8.6 m) sin39 = 5.41 m, and v y = − y ) = 2(9.8 m/s )(5.41 m) = 10.3 m/s 2g(y (see Equation 2-11). The time to reach the apex is t = v y /g, so v x = (x − x )/t = (x − x )g/v y (see Equations 4-6 and 4-7). The horizontal distance from apex to origin is x − x = (8.6 m)cos39 = 6.68 m, so ymax
◦
0
max
0
0
0
2
0
0
0
0
◦
0 2
v0x = (6.68 m)(9.8 m/s )/(10.3 m/s) = 6.36 m/s. v 0 can be expressed in unit vector notation as (6.36ˆı+ 10.3ˆ) m/s, or by its magnitude v02x + v02y = 12.1 m/s and direction θ = tan (CCW from the x-axis).
1
−
(v0y /v0x ) = 58.3
◦
ac = 21 [(390/3.6)2 + (740/3.6)2](m2 /s2 )/(7.1 km) = 3.80 m/s2 . Finally, θ = tan 1 (3.80/0.456) = 83.2 . (Instead of working out each component of acceleration numerically, we could have written the final result symbolically as follows: −
ac = at
vi2 + vf 2 2r
and θ = tan before.)
1
−
◦
3π · 2(3/4)(2πr) = (v − v ) 2 2
f
2
i
[3π(7402 + 3902 )/2(7402
vf 2 + vi2 vf 2 vi2
−
,
2
− 390 )] as
Problem 66. On its landing approach a plane makes a semicircular turn of radius 6.7 km, remaining at constant altitude while its speed drops steadily from 840 km/h to 290 km/h. At the midpoint of the turn, what is the angle between the plane’s velocity and acceleration vectors?
Problem 64. A circus lion prepares to leap through a flaming hoop. A line from the lion to the hoop is 2.2 m long and makes a 26 angle with the floor. With what initial velocity should the lion leap so as to pass through the hoop moving horizontally? ◦
Solution
Solution The apex of the lion’s trajectory (where it’s moving horizontally with v x = v 0x and v y = 0) is displaced from the origin from which it leaps by x x0 = (2.2 m)cos26 = 1.98 m, and y y0 = (2.2 m)sin26 = 0.964 m. As in the previous problem, v0y = 2g(y y0 ) = 4.35 m/s, and v 0x = (x x0 )g/v 0y = 4.46 m/s. The magnitude and direction of v 0 are 6.23 m/s and 44.3 (CCW from x-axis). ◦
−
−
◦
−
35
−
The symbolic result of the previous problem can be used if we change the length of turn from 43 to 21 of a circle. Then θ = tan 1 [π(vf 2 + vi2 )/(vf 2 vi2 )] = tan 1 [π(2902 + 8402 )/(2902 8402 )] = 104 . Since the plane decelerates, the argument of the arctan is negative, and the angle is in the second quadrant (i.e., at is opposite to the velocity). −
−
−
−
◦
◦
Supplementary Problems Problem
Problem 65. After takeoff, a plane makes a three-quarter circle turn of radius 7.1 km, maintaining constant altitude but steadily increasing its speed from 390 to 740 km/h. Midway through the turn, what is the angle between the plane’s velocity and acceleration vectors?
Solution
67. Verify the maximum altitude and flight time for the 15 launch angle trajectory of the missile described at the end of the Application: Ballistic Missile Defense (page 79). ◦
Solution The missile has a range (on level ground) of R = xmax x0 = 1000 km at a launch angle of θ0 = 15 , so Equation 4-10 gives the launch speed as v02 = gR/ sin2θ0 . (Numerically, this is ◦
The plane’s velocity is tangent to its circular path in the direction of motion and so is the tangential acceleration a t . The radial (centripetal) acceleration ac is perpendicular to this, so the angle between the total acceleration and the velocity is θ = tan 1 (ac /at ), inclined toward the center of the turn. For the linear motion over a circular distance of s = 43 (2πr) = 1.5π(7.1 km) = 33.5 km, Equation 2-11 can be used to 2 find the constant a t , at = (vf vi2 )/2s = [(740/3.6)2 (390/3.6)2](m2 /s2 )/2(33.5 km) = 0.456 m/s2 . We can find a c = v 2 /r midway through the turn by using Equation 2-11 again, since v 2 = v i2 + 2at (s/2) = 2 2 vi2 + 21 (vf vi2 ) = 21 (vi2 + vf ). Then −
−
−
−
−
(0.0098 km/s2 )(1000 km)/ sin2(15 ) = 4.43 km/s, as stated in the text.) The maximum altitude can be found from Equation 2-11, h = y max y0 = v 02y /2g, since v y = 0 at this point in the trajectory. With the value of v 0 above, we find h = v 02 sin2 θ0 /2g = (gR/ sin2θ0 )sin2 θ0 /2g = R sin2 θ0 /2sin2θ0 = 1 R tan θ0 = 41 (1000 km) tan15 = 67.0 km, in 4 agreement with the text. (To answer just this question, one might have substituted numbers into the first expression for h , i.e., (4.43 km/s)2 sin2 15 /2g, but working algebraically, the simpler expression from ◦
−
◦
◦
36
CHAPTER 4
Problem 33 is obtained. Moreover, in this particular case, 2 sin 2θ0 = 2sin2(15 ) = 1, so the numerical calculation is faster too.) The time of flight can be found from Equation 4-7, t = (xmax x0 )/v0x = R/v0 cos θ0 = (1000 km)/(4.43 km/s) cos15 = 234 s = 3.90 min, completing the verification of the text. (Again, alternate expressions could have been used, e.g., t = 2v0y /g, from Equation 4-6, or t = R/2g (cos15 ) 1 .) ◦
−
◦
◦
−
Problem 68. A juggler’s hands are 80 cm apart, and the balls being juggled reach a maximum height of 100 cm above the juggler’s hands. (a) At what velocity do the balls leave the juggler’s hands? (b) If four balls are being juggled, how often must the juggler catch a ball?
Equation 2-11, evaluated at the highest point, v02y = v02 sin2 θ0 = 2gy max. From Equation 4-10, v02 sin θ0 cos θ0 = 21 gx max. Dividing these, we find tan θ0 = 4ymax/xmax = 4(100 cm)/(80 cm) = 5, or θ0 = tan 1 5 = 78.7 . Substituting this angle into the first equation (for example), we find v 0 = −
◦
2(9.8 m/s2 )(1 m)/ sin78.7 = 4.51 m/s. (b) The time of flight for one of the balls can be calculated from Equation 4-8 (with y = y0 = 0) and the answer to part (a) above. Thus, t = 2v0y /g = 2 2gy max/g = ◦
√
2 2ymax/g = 2 2(1 m)/(9.8 m/s2 ) = 0.904 s. Suppose the juggler throws a ball with the left hand while simultaneously catching another with the right. At this instant, the other two balls occupy positions in flight along the trajectory. During the time of flight of a given ball, three balls must be caught, namely, the given ball and the two already in the air. Thus, three catches must be made in 0.904 s, or one catch every 0.301 s. (This is also the time during which a ball must be transferred from the catching (right) hand to the throwing (left) hand. Thus, the total time for a given ball to circulate once around completely would be 0.904 s + 0.301 s 1.21 s.)
≈
Problem 69. A monkey is hanging from a branch a height h above the ground. A naturalist stands a horizontal distance d from a point directly below the monkey. The naturalist aims a tranquilizer dart directly at the monkey, but just as she fires the monkey lets go. Show that the dart will nevertheless hit the monkey, provided its initial speed exceeds (d2 + h2 )g/2h.
Solution Gravity accelerates the dart and the monkey equally, so both fall the same vertical distance from the point of aim (the monkey’s original position) resulting in a hit, provided the initial speed of the dart is sufficient to reach the monkey before the monkey reaches the ground. To prove this assertion, let the dart be fired from ground level (y = 0) with speed v 0 and direction θ0 = tan 1 (h/d) (line of sight from naturalist N to monkey M ), while the monkey drops from height h at t = 0. The vertical height of each is y monkey = h 21 gt 2 and y dart = v 0y t 21 gt2 , where v 0y = v 0 sin θ0 = v0 h/ d2 + h2 . (The term 21 gt2 represents the effect of gravity, which appears the same way in both y -coordinate equations.) The dart strikes the monkey when y monkey = ydart , which implies h = v 0y t, or t = d2 + h2 /v0 . This must be less than the time required for the monkey to fall to the ground, which is 2h/g (from y monkey = 0). Thus −
Problem 68 Solution.
Solution (a) Suppose the juggler’s hands are located at the origin and 80 cm along the x-axis, as shown. From
−
√
√
−
−
CHAPTER 4
√ d
2
+ h2 /v0 < 2h/g or v 0 > g(d2 + h2 )/2h. (This condition can also be understood from the horizontal range formula, Equation 4-10. The range of the dart has to be greater than the horizontal distance to the monkey, d < v 02 sin 2θ0 /g = v 02 2hd/(d2 + h2 )g.)
37
θ0 = tan 1 a, and v 0 = g(1 + a2 )/2b, so with the given numerical values, one finds −
7.4a = and 7.4b = or θ 0 = tan
1
−
9.5 2.1
2.1 9.5
−
1 2.1 m
−
3.2,
1 9.5 m
3.2,
1.86 = 61.7 , and v 0 = ◦
(9.8 m/s2 )(1 + 1.862)/2(0.160 m
1)
−
= 11.7 m/s.
Problem 71. A diver leaves a 3-m board on a trajectory that takes her 2.5 m above the board, and then into the water a horizontal distance of 2.8 m from the end of the board. At what speed and angle did she leave the board?
Problem 69 Solution.
Problem 70. A child tosses a ball over a flat-roofed house 3.2 m high and 7.4 m wide, so it just clears the corners on both sides, as shown in Fig. 4-37. If the child stands 2.1 m from the wall, what are the ball’s initial speed and launch angle? Assume the ball is launched essentially from ground level. y
( x2, y2)
( x1, y1)
Problem 71 Solution. 3.2 m x
( x0, y0)
7.4 m
2.1 m figure
( x3, y3)
4-37 Problem 70 Solution.
Solution Choose the coordinate system shown in Fig. 4-37, such that the four given points on the trajectory are the origin (x0 , y0 ) = (0, 0), the first corner (x1 , y1 ) = (2.1 m, 3.2 m), the second corner (x2 , y2 ) = (9.5 m, 3.2 m), and ground level at the right (x3 , y3 ) = (11.6 m, 0). Equation 4-9 for the trajectory, evaluated at any two points other than the origin, provides two equations which can be solved for v 0 and θ0 . Before substituting values, we may divide Equation 4-9 by x, and let a = tan θ0 and b = g/2v02 cos2 θ0 . For example, selecting the two corner points, we obtain y1 /x1 = a x1 b, and y 2 /x2 = a x2 b, with solutions (x2 x1 )a = (x2 y1 /x1 ) (x1 y2 /x2 ), and (x2 x1 )b = (y1 /x1 ) (y2 /x2 ). In terms of a and b ,
− −
−
−
−
−
Solution Since we are given the maximum height (at which point v y = 0), Equation 2-11 can be used to find the y component of the diver’s initial velocity, 0 = v 02y 2g(ymax y0 ) or v 0y =
−
−
2(9.8 m/s2 )(2.5 m) = 7.00 m/s. (We take the positive square root because the diver springs upward off the board.) The x-component of v0 can be found from Equation 4-7, once the time of flight is known. The latter is the positive root (the dive begins at t = 0) of the quadratic Equation 4-8, when y 0 y = 3 m (a 3-m board is 3 m above the water level).
−
Thus, t = [v0y +
v02y + 2g(y0
− y)]/g = [7 m/s +
49 m2 /s2 + 2(9.8 m/s2 )(3 m)]/(9.8 m/s2 ) = 1.77 s, and v 0x = (x x0 )/t = 2.8 m/1.77 s = 1.58 m/s. From v0x and v 0y we find the magnitude v0 =
−
v02x + v02y = 7.18 m/s and direction
θ0 = tan
1
−
(v0y /v0x ) = 77.3 . ◦
38
CHAPTER 4
72. In your calculus class, you may have learned that you can find the maximum or minimum of a function by differentiating and setting the result to zero. Do this for Equation 4-10, differentiating with respect to θ 0 , and thus verify that the maximum range occurs for θ 0 = 45 . ◦
The derivative of Equation 4-10 with respect to θ 0 is dx/dθ0 = 2(v02 /g)cos2θ0 . This is zero when 2θ0 = 90 , or θ 0 = 45 as stated. (This is the only maximum, since launch angles are restricted to the range 0 < θ0 < 90 , and d 2 x/dθ02 < 0.) ◦
◦
◦
Problem 73. A projectile is launched with speed v 0 from the edge of a cliff of height h ; the ground below the cliff is flat. Using the technique of the preceding problem, show that the maximum range occurs when the launch angle is v0 θ0 = tan 1 . 2gh + v02 −
Solution
The quadratic formula can be used to solve Equation 4-9 for the horizontal range of a projectile with positive v 0x , whose trajectory begins at the origin and ends at the point (x, y) : x = (v02 cos2 θ0 /g)[tan θ0 tan2 θ0 2gy/v02 cos2 θ0 ]. When the origin is on a cliff of height h above where the projectile lands, y = h, and only the solution with the positive sign before the square root corresponds to a positive range. (For y > 0, both solutions might be possible.) Thus, the horizontal range appropriate to the situation in this problem is
±
−
−
x = (v02 cos θ0 /g)[sin θ0 + sin2 θ0 + 2gh/v02 ]. We chose to multiply through one factor of cos θ0 in order to simplify the θ 0 dependence of each term before differentiating; other choices also work. Inspection of the trajectory shows that it is reasonable to expect one maximum value of x, with v 0 constant, for 0 < θ0 < 90 . It can be found by the method suggested, or by other methods, such as Lagrange multipliers. In taking the derivative, we use the Product rule, the Chain rule, and the derivatives of the sine, cosine, and square root ( z = z 1/2 ) given in Appendix A. Then,
−
0
0
0
= 0,
√ where · ·· = sin θ + 2gh/v . The first factor is √ never zero, so · · · = cos θ / sin θ = (1/ sin θ ) − sin θ . Squaring and simplifying, we find
0
2
2 0
0
2
0
0
0
that
−
2gh 1 1 = sin θ0 + 2 = 2 (sin θ0 ) 2 v0 sin θ0 sin θ0 2gh +sin2 θ0 , or 2 + 2 v0 1 1 = =1 + , 2 tan2 θ0 sin θ0 v0 or tan θ0 = , as stated. (With this value 2 v0 + 2gh for θ 0 , the maximum value of x turns out to be (v02 /g) 1 + 2gh/v02 . For h = 0, these values of θ 0 and x reduce to the case discussed in the text.)
√ ( . . .)
Solution
2
√ cos θ = [sin θ + · · ·] − sin θ + √ ···
Problem
2
2
Problem 74. Two projectiles are launched with the same speed v0 , at angles 45 + α and 45 α. As Fig. 4-15 shows, they have the same horizontal range. Derive an expression for the difference in their flight times. ◦
◦
−
Solution The time of flight for a parabolic trajectory between points at the same altitude (y y0 = 0) is t = 2v0y /g (see Equation 4-6 with v y = v0y , or the solution to Problem 67). The difference in flight-times for two trajectories with the same horizontal range and launch speeds, but different angles, is
− −
◦
◦
∆t = (2v0 /g)[sin(45 + α) sin(45 α)] = (2v0 /g)2 cos 12 (45 + α + 45 α) 1 sin 2 (45 + α 45 + α) ◦
×
◦
◦
−
−
◦
◦
− −
√
= (4v0 /g)cos45 sin α = (2 2v0 /g)sin α. (We used an identity for the difference of sines from Appendix A.)
◦
√
g dx = [sin θ0 + v02 dθ0
√ . . .](− sin θ ) 0
+cos θ0 cos θ0 +
sin θ0 cos θ0 ...
√
Problem 75. A well-engineered ski jump is less dangerous than it looks because skiers hit the ground with very small velocity components perpendicular to the ground. Skiers leave the Olympic ski jump in Lake Placid, New York, at an angle of 9.5 below the horizontal. Their landing zone is a horizontal distance of 55 m from the end of the jump. The ground at the point is contoured so skiers’ trajectories make an angle of only 3.0 with the ◦
◦
CHAPTER 4
39
Solution
ground on landing, as suggested in Fig. 4-38. What is the slope of the ground in the landing zone?
Suppose the particle passes the origin at time t = 0. Then x = v 0 t, y = 21 at2 , vx = v 0 , and v y = at. The tangents of the angles made by the velocity and displacement vectors with the x-axis are v y /vx = at/v0 and y /x = 21 at2 /v0 t = 21 at/v0 = 21 vy /vx , respectively, as asserted.
55 m °
9.5
Problem 77. Derive a general expression for the flight time of a projectile launched on level ground with speed v 0 and launch angle θ 0 . 3
figure
°
Equation 4-8 gives the height of a projectile above its launch site, y y0 = (v0y 21 gt)t. Here, we factored the time to show that the two solutions for zero height are the launch time t = 0, and the time of flight, given by v 0y 21 gt = 0, or t = 2v0y /g = 2v0 sin θ0 /g.
−
4-38 Problem 75.
The direction of the skier’s velocity is θ = tan 1 (vy /vx ), where angles are measured CCW from the x-axis, chosen horizontal to the right in Fig. 4-38 with the y -axis upward. In the landing zone, θ is in the fourth quadrant, which can be represented by a negative angle below the x-axis. The slope of the ground at this point can be represented by a similar angle θ g , and for the safety of ski jumpers, θg θ = 3.0 . The slope v y /vx = (dy/dt)/(dx/dt) = dy/dx can be calculated by differentiating the trajectory equation, but it is just as easy in this problem to use Equations 4-5, 4-6 and 4-7. Thus, v x = v 0x = v 0 cos θ0 , and v y = v 0y gt = v 0 sin θ0 gt. The time of flight can be eliminated, since x x0 = v 0x t = 55 m is given, so v y /vx = (v0y /v0x ) g(x x0 )/v02x = tan θ0 g(x x0 )/v02 cos2 θ0 = tan( 9.5 ) (9.8 m/s2 )(55 m)/(28 cos( 9.5 ) m/s)2 = 0.874. Finally, θ = tan 1 ( 0.874) = 41.2 , and θg = θ + 3.0 = 38.2 . −
◦
−
−
−
−
Solution
−
Solution
−
◦
−
◦
− − − − − − − −
−
◦
◦
Problem 78. An object moves at constant speed v in the x-y plane, describing a circle of radius r centered at the origin. It is on the positive x-axis at time t = 0. Show that the position of the object as a function of time can be written r = r[cos(vt/r)ˆı + sin(vt/r)ˆ], where the argument of the sine and cosine is in radians. Differentiate this expression once to obtain an expression for the velocity and again for the acceleration. Show that the acceleration has magnitude v 2 /r and is directed radially inward (that is, opposite to r).
− −
◦
Problem 76. A particle is moving along the x-axis with velocity v0 in the positive x direction. As it passes the origin, it begins to experience a constant acceleration a in the y direction. Show that, at any subsequent time, the tangent of the angle its velocity makes with the x-axis is twice the tangent of the angle its displacement vector makes with the x-axis.
Problem 78 Solution.
Solution From the diagram, r = r(cos θˆı+ sin θˆ). In radians, θ = s/r, where s is the arc-length, and for constant speed, s = vt. Thus, r = r[cos(vt/r)ˆı + sin(vt/r)ˆ].
40
CHAPTER 4
d The velocity v = dr/dt. Since dt (sin ωt) = ω cos ωt, d ω sin ωt, where ω (omega) = v/r is dt (cos ωt) = constant, we have v = v[ sin(vt/r)ˆı + cos(vt/r)ˆ]. (Note that v r, i.e., v is tangent to the circle.) The acceleration a = dv/dt = (v2 /r)[cos(vt/r)ˆı + sin(vt/r)ˆ] has magnitude v 2 /r and direction opposite to r, i.e., toward the center of the circle. (This kind of acceleration is therefore called a centripetal, or “center-seeking,” acceleration.)
− ⊥
−
−
defined as R = [1+ (dy/dx)2 ]3/2 /(d2 y/dx2 ), where the positive (negative) sign is used if the arc length increases in the positive (negative) x direction. (See any comprehensive calculus text.) For the trajectory of Equation 4-9, dy/dx = tan θ0 x(g/v 02 cos2 θ0 ), d2 y/dx2 = g/v 02 cos2 θ0 , and the positive sign in the expression for R is used when cos θ0 0. Of course, dy/dx = v y /vx gives the direction of the instantaneous velocity, which is zero at the apex of the trajectory. Then R = (d2 y/dx2 ) 1 = v02 cos2 θ02 /g = vx2 /g at the apex, as above. In terms of the horizontal range, the radius of curvature at an arbitrary point x (0 x xmax) on the trajectory for which Equation 4-10 applies is R = [1 + (1 2x/xmax)2 tan2 θ0 ]3/2 ( xmax/2tan θ0 ), since dy/dx = (1 2x/xmax)tan θ0 , and d 2 y/dx2 = 2tan θ0 /xmax.
±
−
−
≥
−
Problem 79. In the Olympic hammer throw, contestants whirl a 7.3-kg ball on the end of a 1.2-m-long steel wire before releasing it. In a particular throw, the hammer is released from a height of 1.3 m while moving in a direction 24 above the horizontal. If it travels 84 m horizontally, what is its radial acceleration just before release (see Fig. 4-39)? ◦
Solution Just before release, the hammer ball is traveling in a circle (approximately of radius r = 1.2 m), so its radial (or centripetal) acceleration is a c = v 02 /r, where v 0 is the launch speed. We can determine v 0 from Equation 4-9 and the data given for the throw, θ0 = 24 , x = 84 m, y = 1.3 m, as in Example 4-5, thus: v2 1 gx 2 ac = 0 = = 892 m/s2 . r r 2cos2 θ0 (x tan θ0 y) ◦
−
−
− −
−
−
×
Problem 81. Two golfers stand equal distances on opposite sides of a hole, as shown in Fig. 4-40. Golfer A hits his ball at a 50 angle to the horizontal. At the instant she hears A’s club hit the ball, Golfer B hits her ball at the same speed as A, but at a 40 angle. If the two balls reach the hole simultaneously, how far apart are the golfers? The speed of sound is 340 m/s. ◦
◦
900
Problem
figure
Solution
−
≤ ≤
80. A projectile is launched at an angle θ 0 to the horizontal, with sufficient speed to give it a horizontal range x. Show that the radius of curvature at the top of its trajectory is given by r = x/2tan θ0 .
−
4.40 Problem 81 Solution
Solution Since both golfers and the hole are on level ground, we can use the result of Problem 77 to determine their separation, R. If golfer A’s ball is hit at t = 0, it will reach the hole at t = 2v0 sin 50 /g. Golfer B’s ball has a time of flight of 2v0 sin 40 /g, but it starts after a delay of R/(340 m/s), due to the sound travel-time. Since both balls arrive at the hole simultaneously (a double hole-in-one!), 2v0 sin40 /g + R/(340 m/s) = 2v0 sin 50 /g, or R = (340 m/s)(2v0 /g) (sin 50 sin 40 ). We can eliminate v 0 by using Equation 4-10, with range R/2, since the launch angles are complementary. Then, R/2 = (v02 /g)sin2θ0 = (2v02 /g)sin40 sin 50 , or 2v0 = gR/ sin40 sin 50 . Substituting above, we find R = (340 m/s) ( gR/ sin40 sin50 /g)(sin50 sin40 ), or (340 m/s)2 (sin 50 sin40 )2 R = = 364 m. (9.8 m/s2 )sin40 sin50 ◦
At the apex of a parabolic trajectory, the acceleration is perpendicular to the velocity (v = v x , vy = 0) and is therefore entirely radial (a = ar , at = 0). In projectile motion, the acceleration has constant magnitude g , so at the apex, the radius of curvature has magnitude R = v 2 /ar = vx2 /g = (v02 /g)cos2 θ0 . In terms of the horizontal range, Equation 4-10, this becomes R = (xmax/ sin2θ0 )cos2 θ0 = xmax /2tan θ0 . Actually, a r = g, and the radius of curvature is negative, R = vx2 /g at the apex. It is only the magnitude of R that is discussed in the text and specified in this problem. A negative R means that the slope of the tangent decreases as the trajectory is traced in the direction of increasing arc length. For a plane curve y = f (x), the radius of curvature is
−
−
◦
◦
◦
◦
×
◦
−
◦
◦
◦
◦
◦
× ◦
◦
◦
− ◦
−
◦
◦
◦
CHAPTER 4
Problem 82. A convertible is speeding down the highway at 130 km/h, when the driver spots a police airplane 600 m back at an altitude of 250 m. The driver decelerates at 2.0 km/h/s. (a) If the plane is flying horizontally at a steady 210 km/h, where should the plane be in relation to the car for the police officer to drop a speeding ticket into the car? Assume the ticket is dropped with no initial vertical motion, and it is in a heavy capsule that experiences negligible air resistance. (b) How fast will the car be moving when the ticket reaches it?
Problem 82 Solution.
Solution (a) Suppose that the road is horizontal and straight, and that the velocities of the car and plane are collinear. In the spirit of Section 3-5, let us work in a reference frame S attached to the car. At t = 0 (when the car spots the plane) S is 600 m from the origin of a frame S attached to the road. The initial horizontal velocity of the capsule relative to S is v 0 = v 0 V 0 = 210 130 = 80 km/h. The horizontal acceleration of the capsule relative to S is a x = ax acar = 0 ( 2 km/h/s) = 2 km/h/s. (S is an accelerated frame, so we used the derivative of Equation 3-10 with dV/dt = a car.) The horizontal position of the capsule relative to S is therefore x (t) = x0 + v0 t + 21 ax t2 , where x 0 = 600 m. The vertical motion of the capsule is the same in S and S , so if the capsule is dropped at time t d , its height is y = 250 m for 0 t td , and y = 250 m 21 g(t td )2 for t d t. The capsule arrives at time t a , when y = 0, so t a td = 2y0 /g = ′
′
′
− −−
′
′
′
−
−
′
′
′
′
′
′
′
−
′
−
≤ ≤ ≤ −
−
2(250 m)/(9.8 m/s2 ) = 7.14 s. When the capsule arrives, x (ta ) = 0 (it’s at the car), so 0 = x0 + v0 ta + 21 ax t2a = ( 0.6 km) (3600 s/h) + ta (80 km/h)+ 21 t2a (2 km/h/s), or 0 = 2160 s2 + ta 80 s + t2a . Taking the positive solution of this quadratic, we find t a = 402 + 2160 40 = 21.3 s. Therefore, t d = 21.3 s 7.14 s = 14.2 s, and x (td ) = 600 m + (80 km/h)(14.2 s)+ 1 (2 km/h/s)(14.2 s)2 = 229 m. (The plane should be 2 229 m behind the car when the capsule is dropped.) (b) Finally, the velocity of the car, relative to the ′
′
′
′
−
′
−
−
√ −
−
×
−
41
ground, at t a is V = V 0 + acar ta = 130 km/h + ( 2 km/h/s)(21.3 s) = 87.4 km/h = 54.3 mi/h, just under the urban highway speed limit.
−
Problem 83. A projectile is launched with initial speed v 0 at an angle θ 0 to the horizontal. Find expressions for
42
CHAPTER 4
the angle the trajectory makes with the horizontal (a) as a function of time and (b) as a function of position.
Solution (a) The slope of the trajectory is v y /vx , and the angle it makes with the x-axis is θ = tan 1 (vy /vx ). For projectile motion, v x = v 0x = v 0 cos θ0 is a constant, and v y = v 0y gt = v 0 sin θ0 gt; therefore, v y /vx = (v0y /v0x ) (gt/v0x ) = tan θ0 gt/v0 cos θ0 . (b) We can eliminate t from the expression for the slope by using x x0 = v 0x t, or t = (x x0 )/v0x . Thus, vy /vx = tan θ0 g(x x0 )/v02 cos2 θ0 . (The expression for the angles is the inverse tangent of the slopes above.) −
−
−
−
−
−
− − −
