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
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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)
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
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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
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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=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!)
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!)
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
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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. answer
Solution in SI Units
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
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
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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
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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)
System International Solution
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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)
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
625=12(0.03125)t2 t=200 sec
t=3 min 20 sec answer
0.3048=12a(82) a=0.009525 m/sec2
190.5=12(0.009525)t2 t=200 sec
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.
Solution in English Units
HideClick here to show or hide the solution vf=vi+at 24=vo+10a → equation (1) 18=vo+12a → equation (2)
Equation (1) minus equation (2)
6=−2a
a=−3 ft/sec2 answer From equation (1) 24=vo+10(−3) vo=54 ft/sec answer Solution in SI Units
HideClick here to show or hide the solution vf=vi+at 7.32=vo+10a → equation (1) 5.49=vo+12a → equation (2)
Equation (1) minus equation (2)
1.83=−2a
a=−0.915 m/sec2 answer
From equation (1)
7.32=vo+10(−0.915)
