The best way to escape from a problem is to solve it.This ebook helps in your day to day problem solving by using very simple problem solving techniques.
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Problem 1003 A stone is thrown vertically upward and return to earth in 10 sec. What was its initial velocity and how high did it go?
Solution
HideClick here to show or hide the solution Return in 10 seconds = 5 seconds upward + 5 seconds downward SI Units Going upward (velocity at the highest point is zero):
vf=vi−gt 0=vi−9.81(5) vi=49.05 m/sec
answer
Going downward (initial velocity is zero; free-fall):
h=12gt2 h=12(9.81)(52) h=122.625 m
answer
English System Going upward (velocity at the highest point is zero):
vf=vi−gt 0=vi−32.2(5) vi=161 ft/sec
answer
Going downward (initial velocity is zero; free-fall):
h=12gt2 h=12(32.2)(52) h=402.5 ft answer
Problem 1004 A ball is dropped from the top of a tower 80 ft (24.38 m) high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec (12.19 m/s). When and where do they pass, and with what relative velocity?
Solution in English System
HideClick here to show or hide the solution h = 80 ft vA = 0 vB = 40 ft/s g = 32.2 ft/s2 From A to C (free-fall)
h1=12gt2 h1=12(32.2)t2 h1=16.1t2 From B to C (upward motion) From the formula s = vit + ½ at2
h2=vBt−12gt2 h2=40t−12(32.2)t2 h2=40t−16.1t2 A to C plus B to C is equal to height of the tower
h1+h2=h 16.1t2+(40t−16.1t2)=80 40t=80 t=2 sec h1=16.1(22) h1=64.4 ft They pass each other after 2 seconds at 64.4 ft from the top of the tower.
answer
Velocity at C of stone from A (after 2 seconds)
vC1=gt=32.2(2) vC1=64.4 ft/s Velocity at C of stone from B (after 2 seconds)
vC2=vB−gt==40−32.2(2) vC2=−24.4 ft/s → the negative sign indicates that the stone is moving downward
Relative velocity:
vr=vC1+vC2=64.4−24.4 vr=40 ft/sec answer Solution in SI Units
HideClick here to show or hide the solution h = 24.38 m vA = 0 vB = 12.19 m/s g = 9.81 m/s2 From A to C (free-fall)
h1=12gt2 h1=12(9.81)t2 h1=4.905t2 From B to C (upward motion) From the formula s = vit + ½ at2
h2=vBt−12gt2 h2=12.19t−12(9.81)t2 h2=12.19t−4.905t2 A to C plus B to C is equal to height of the tower
h1+h2=h 4.905t2+(12.19t−4.905t2)=24.38 12.19t=24.38 t=2 sec h1=4.905(22) h1=19.62 m They pass each other after 2 seconds at 19.62 m from the top of the tower.
answer
Velocity at C of stone from A (after 2 seconds)
vC1=gt=9.81(2) vC1=19.62 m/s Velocity at C of stone from B (after 2 seconds)
vC2=vB−gt==12.19−9.81(2) vC2=−7.43 m/s → the negative sign indicates that the stone is now moving downward Relative velocity:
vr=vC1+vC2=19.62−7.43
vr=12.19 m/sec
answer
Problem 1005 A stone is dropped down a well and 5 sec later, the sounds of the splash is heard. If the velocity of sound is 1120 ft/sec (341.376 m/s), what is the depth of the well?
Solution
HideClick here to show or hide the solution
tstone+tsound=5 For tstone (free-falling body):
h=12gt2 t=2hg−−−√ For tsound (uniform motion):
h=vt t=hv Thus,
2hg−−−√+hv=5 English System
2h32.2−−−−√+h1120=5 2h32.2−−−−√=5−h1120 2h32.2=(5−h1120)2 10h161=25−h112+h21254400 11254400h2−1832576h+25=0 h=88759.73 and 353.31 For h = 88 759.73 ft
t=2(88759.73)32.2−−−−−−−−−−√ t=74.2 sec >5 sec (not okay!) For h = 353.31 ft
t=2(353.31)32.2−−−−−−−−√ $t = 4.68 \, \text{ sec } okay!) Thus, h = 353.31 ft
→ answer
SI Units
2h9.81−−−−√+h341.376=5 2h9.81−−−−√=5−h341.376 200h981−−−−−√=5−125h42672 200h981=(5−125h42672)2 200h981=25−625h21336+15625h21820899584 156251820899584h2−16267756976872h+25=0 h=27065.05 and 107.64 For h = 27 065.05 m
t=2(27065.05)9.81−−−−−−−−−−√ t=74.2 sec >5 sec (not okay!) For h = 107.64 m
t=2(107.64)9.81−−−−−−−−√ $t = 4.68 \, \text{ sec } okay!) Thus, h = 107.64 m
→ answer
Problem 1007 A stone is dropped from a captive balloon at an elevation of 1000 ft (304.8 m). Two seconds later another stone is thrown vertically upward from the ground with a velocity of 248 ft/s (75.6 m/s). If g = 32 ft/s2 (9.75 m/s2), when and where the stones pass each other?
Solution in English Units
HideClick here to show or hide the solution
h=1000 ft vi2=248 ft/s g=32 ft/s2 Stone dropped from captive balloon (free-falling body):
h1=12gt2=12(32)t12 h1=16t12 Stone thrown vertically from the ground 2 seconds later
s=vi2t2−12gt22 h2=248t2−12(32)t22 h2=248(t1−2)−16(t1−2)2 h1+h2=h 16t12+[248(t1−2)−16(t1−2)2]=1000 16t12+248(t1−2)−16(t12−4t1+4)=1000 16t12+248t1−496−16t12+64t1−64=1000 16t12+248t1−496−16t12+64t1−64=1000 312t1=1560 t1=5 sec The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon. answer
h2=248(5−2)−16(5−2)2 h2=600 ft The stones will meet at a point 600 ft above the ground.
Solution in SI Units
HideClick here to show or hide the solution
answer
h=304.8 m vi2=75.6 m/s g=9.75 ft/s2 Stone dropped from captive balloon (free-falling body):
h1=12gt2=12(9.75)t12 h1=4.875t12 Stone thrown vertically from the ground 2 seconds later
s=vi2t2−12gt22 h2=75.6t2−12(9.75)t22 h2=75.6(t1−2)−4.875(t1−2)2 h1+h2=h 4.875t12+[75.6(t1−2)−4.875(t1−2)2]=304.8 4.875t12+75.6(t1−2)−4.875(t12−4t1+4)=304.8 4.875t12+75.6t1−151.2−4.875t12+19.5t1−19.5=304.8 4.875t12+75.6t1−151.2−4.875t12+19.5t1−19.5=304.8 95.1t1=475.5 t1=5 sec The stones will pass each other 5 seconds after the first stone was dropped from the captive balloon. answer
h2=75.6(5−2)−4.875(5−2)2 h2=182.925 m The stones will meet at a point 182.925 m above the ground.
answer
Problem 1008 A stone is thrown vertically upward from the ground with a velocity of 48.3 ft per sec (14.72 m per sec). One second later another stone is thrown vertically upward with a velocity of 96.6 ft per sec (29.44 m per sec). How far above the ground will the stones be at the same level?
Solution in English System
HideClick here to show or hide the solution
s=vit+12at2 h=vit−12gt2 h=vit−12(32.2)t2 h=vit−16.1t2 For the first stone:
h1=48.3t−16.1t2 For the second stone
h2=96.6(t−1)−16.1(t−1)2 h2=96.6(t−1)−16.1(t2−2t+1) h2=96.6t−96.6−16.1t2+32.2t−16.1 h2=−16.1t2+128.8t−112.7 h1=h2 48.3t−16.1t2=−16.1t2+128.8t−112.7 80.5t=112.7 t=1.4 s h=48.3(1.4)−16.1(1.42) h=36.064 ft answer Solution in SI
HideClick here to show or hide the solution
s=vit+12at2 h=vit−12gt2 h=vit−12(9.81)t2 h=vit−4.905t2 For the first stone:
h1=14.72t−4.905t2
For the second stone
h2=29.44(t−1)−4.905(t−1)2 h2=29.44(t−1)−4.905(t2−2t+1) h2=29.44t−29.44−4.905t2+9.81t−4.905 h2=−4.905t2+39.25t−34.345 h1=h2 14.72t−4.905t2=−4.905t2+39.25t−34.345 24.53t=34.345 t=1.4 s h1=14.72(1.4)−4.905(1.42) h=10.994 m answer
Problem 1009 A ball is shot vertically into the air at a velocity of 193.2 ft per sec (58.9 m per sec). After 4 sec, another ball is shot vertically into the air. What initial velocity must the second ball have in order to meet the first ball 386.4 ft (117.8 m) from the ground?
Solution: English System of Units
HideClick here to show or hide the solution
s=vit+12at2 a=−g=−32.2 ft/s2 s=386.4 ft Thus,
386.4=vit−16.1t2 First ball:
386.4=193.2t–16.1t2 t2−12t+24=0 t=9.46 and 2.5 Use t=9.46 s Second ball:
386.4=vi(t−4)−16.1(t−4)2 386.4=vi(9.46−4)−16.1(9.46−4)2 vi=158.67 ft/s answer Solution: International System of Units
HideClick here to show or hide the solution
s=vit+12at2 a=−g=−9.81 m/s2 s=117.8 ft Thus,
117.8=vit−4.905t2 First ball:
117.8=58.9t–4.905t2 4.905t2−58.9t+117.8=0
t=9.47 and 2.54 Use t=9.47 s Second ball:
117.8=vi(t−4)−4.905(t−4)2 117.8=vi(9.47−4)−4.905(9.47−4)2 vi=48.36 m/s answer
Problem 1010 A stone is thrown vertically up from the ground with a velocity of 300 ft per sec (91.44 m/s). How long must one wait before dropping a second stone from the top of a 600-ft (182.88-m) tower if the two stones are to pass each other 200 ft (60.96 m) from the top of the tower?
English System Solution
HideClick here to show or hide the solution Stone from the ground:
s=vit+12at2 h1=vi1t−12gt2 600−200=300t−1232.2t2 16.1t2−300t+400=0 t=17.19 sec and 1.44 sec Stone from the top of the tower: Let t2 = time to wait before dropping the second stone
h=12g(t−t2)2 With t = 17.19 sec
200=12(32.2)(17.19−t2)2 t2=13.67 sec With t = 1.44 sec
200=12(32.2)(1.44−t2)2 t2=−2.08 sec (meaningless) Use t2=13.67 sec
answer
System International Solution
HideClick here to show or hide the solution Stone from the ground:
s=vit+12at2 h1=vi1t−12gt2 182.88−60.96=91.44t−129.81t2 4.905t2−91.44t+121.92=0 t=17.19 sec and 1.44 sec Stone from the top of the tower: Let t2 = time to wait before dropping the second stone
h=12g(t−t2)2 With t = 17.19 sec
60.96=12(9.81)(17.19−t2)2 t2=13.67 sec With t = 1.44 sec
60.96=12(32.2)(1.44−t2)2 t2=−2.08 sec (meaningless) Use t2=13.67 sec
answer
Problem 1011 A ship being launched slides down the ways with constant acceleration. She takes 8 sec to slide (the first foot | 0.3048 meter). How long will she take to slide down the ways if their length is (625 ft | 190.5 m)?
Solution
HideClick here to show or hide the solution s=vit+12at2 where vi=0 Thus
s=12at2 English Units
SI Units
1=12a(82) a=0.03125 ft/sec2
0.3048=12a(82) a=0.009525 m/sec2
625=12(0.03125)t2 t=200 sec t=3 min 20 sec answer
190.5=12(0.009525)t2 t=200 sec t=3 min 20 sec answer
Problem 1012 A train moving with constant acceleration travels 24 ft (7.32 m) during the 10th sec of its motion and 18 ft (5.49 m) during the 12th sec of its motion. Find its initial velocity and its constant acceleration.
