Question

A golf ball is dropped from rest from a height of 8.10 m. It hits the pavement, then bounces back up, rising just 6.10 m before falling back down again. A boy then catches the ball when it is 1.50 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Answer 1

First find the time taken for the golf ball to hit ground when dropped from 8.1 m

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

since the ball is dropped ,initial velocity u is zero

$s=1/2gt^{2}$

$8.1m=0.5*9.81m/s^{2}*t_{1}^{2}$

$8.1=4.905*t_{1}^{2}$

${\color{Red} t_{1}=1.285s}$

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Now time taken to rise to 6.1 m

First we need initial velocity u for rising part

Since air resistance is ignored,kinetic energy at bottom will be equal to potential energy at height 6.1m

$KE_{bottom}=PE_{6.1m}$

$1/2mu^{2}=mgh$

$u=\sqrt{2gh}$

$u=\sqrt{2*9.81*6.1}$

$u=10.94m/s$

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

Use formula $6.1m=10.94m/s*t_{2}-1/2gt_{2}^{2}$

$6.1=10.94*t_{2}-4.905t_{2}^{2}$

$4.905t_{2}^{2}-10.94*t_{2}+6.1=0$

Solve the quadratic equation using a calculator

${\color{Red} t_{2}=1.11s}$

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Final Part-A boy then catches the ball when it is 1.50 m above the pavement

so height covered =6.1 -1.5 =4.6m

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

since the ball is dropped ,initial velocity u is zero

$s=1/2gt^{2}$

$4.6m=0.5*9.81m/s^{2}*t_{3}^{2}$

$4.6=4.905*t_{3}^{2}$

${\color{Red} t_{3}=0.968s}$

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Total time ,$t=t_{1}+t_{2}+t_{3}$

$t=1.285s+1.11s+0.968s$

ANSWER: ${\color{Red} t=3.363s}$

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