Question

2) Full of hot air. A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sandbag at the instant the balloon is 60.0 m above the ground. After it is released, the sandbag encounters no appreciable air resistance a) Find the position and velocity of the sandbag at 0.200 s and 0.800 s after its release. (6) How much time does it take for the sandbag to strike the ground after it is released? (4) b)

Answer 1

Initial velocity of sandbag,$u_{0}=5m/s$

acceleration,$a=-g=-9.81m/s^{2}$

a)Position

Use Formula $s=ut+1/2at^{2}$

$y=u_{0}t-0.5gt^{2}$

$y=5m/s*0.2s-0.5*9.81*(0.2s)^{2}$

$y=5*0.2-0.5*9.81*0.2^{2}$

$y=0.8038m$

At 0.2s Position is 60.8038 m

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Velocity

Use Formula $v=u+at$

$v=u_{0}-gt$

$v=5m/s-9.81m/s^{2}*0.2s$

ANSWER: ${\color{Red} v_{0.2s}=3.038m/s}$

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Time =0.8s

Use Formula $v=u+at$

$v=u_{0}-gt$

$0m/s=5m/s-9.81m/s^{2}*t$

$t=0.51s$

After $t=0.51s$ sandbag is falling towards earth

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Use Formula $s=ut+1/2at^{2}$

$y=u_{0}t-0.5gt^{2}$

$y=5m/s*0.51s-0.5*9.81*(0.51s)^{2}$

$y=5*0.51-0.5*9.81*0.51^{2}$

$y=1.2742m$

At t=0.51s Max height of 61.2742 m is reached

Time left =0.8-0.51s =0.29s

Position after falling for 0.29s

Use Formula $s=ut+1/2at^{2}$

$y=-0.5gt^{2}$

$y=-0.5*9.81*(0.29s)^{2}$

$y=-0.5*9.81*0.29^{2}$

$y=-0.4125m$

61.2742m-0.4125m=60.8617m

At 0.8s Position is 60.8617m

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Velocity

Use Formula $v=u+at$

$v=u_{0}-gt$

$v=0m/s-9.81m/s^{2}*0.29s$

ANSWER: ${\color{Red} v_{0.8s}=2.8449m/s}$

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b)

Time to hit ground =0.51s +Time taken to fall from Max Height (61.2742m)

Use Formula $s=ut+1/2at^{2}$

$H_{max}=0-0.5gt^{2}$

$61.2742m=0.5*9.81*t^{2}$

$t=12.49s$

Time to hit ground =0.51s + 12.49s

ANSWER: ${\color{Red} t=13s}$

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