Question

A) On Earth (a = 10 m/s/s) you throw a ball downward from the top of a 75 m cliff. The ball takes 3 seconds to strike the ground. How fast was the ball going when you released it?

B) A sports car goes racing down the street and skids to a stop. You measure the length of the skid to be 50 m. This car's brakes give it a 4 m/s/s stopping acceleration. How fast was the car going before braking?

C) You roll a ball up a hill, releasing the ball at 12 m/s. The angle of the hill gives the ball an acceleration of 6.0 m/s/s, directed down the hill. How long will it take the ball to return to you at the bottom of the hill?

Answer 1

A) find initial velocity

time =3s and height =75m

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

$s=u_{y}t+1/2gt^{2}$

$75m=u_{y}*3s+0.5*10m/s^{2}(3s)^{2}$

$75=3u_{y}+0.5*10*3^{2}$

$75=3u_{y}+45$

$30=3u_{y}$

ANSWER: ${\color{Red} u_{y}=10m/s}$

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B)distance covered while braking =50m

deceleration =4m/s²

final velocity =0m/s

Use Formula $v^{2}-u^{2}=2as$

$(0m/s)^{2}-u^{2}=-2*4m/s^{2}*50m$

$0-u^{2}=-2*4*50$

$u=\sqrt{2*4*50}$

ANSWER: ${\color{Red} u=20m/s}$

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C)Use formula

$v=u+at$

Rolling up

Final velocity is zero

so $0m/s=12m/s-6m/s^{2}*t_{1}$

$0=12-6*t_{1}$

$t_{1}=2s$

Use formula $v^{2}-u^{2}=2as$ to find distance covered

$(0m/s)^{2}-(12m/s)^{2}=-2*6m/s^{2}*s$

$0-12^{2}=-2*6*s$

$12^{2}=12s$

$s=12m$

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Rolling down

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

initial velocity is zero

$12m=0+1/2*6m/s^{2}*t_{2}^{2}$

$12=1/2*6*t_{2}^{2}$

$12=3t_{2}^{2}$

$4=t_{2}^{2}$

$t_{2}=2s$

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total time =t₁ +t₂ =2s +2s

ANSWER: ${\color{Red} t=4s}$

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