Question

A golf ball is dropped from rest from a height of 9.50m. It hits the pavement, then bounces back up, rising just 5.70m before falling back down again. A boy then catches the ball on the way down when it is 1.20m 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

Answer:

Explanation:

Given

height from which ball is dropped $h_1=9.5\ m$

and it rises to a height of $h_2=5.7\ m$

A person caught the ball in mid-way i.e. at a height of $h_0=1.2\ m$ from pavement

time taken to reach the bottom is given by

$y=ut+\frac{1}{2}at^2$

where  y=displacement

u=initial velocity

a=acceleration

t=time

here initial velocity is zero i.e.

$h_1=0+\frac{1}{2}gt_1^2$

$t_1=\sqrt{\frac{2h_1}{g}}$

Similarly to reach of  height $h_2$ time taken is $t_2$

$t_2=\sqrt{\frac{2h_2}{g}}$

The ball is caught in mid-way i.e. ball travel a distance of $h_3=5.7-1.2=4.5\ m$

time taken is $t_3=\sqrt{\frac{2h_3}{g}}$

total time taken $t=t_1+t_2+t_3$

$t=\sqrt{\frac{2h_1}{g}}+\sqrt{\frac{2h_2}{g}}+\sqrt{\frac{2h_3}{g}}$

$t=\sqrt{\frac{2\cdot 9.5}{9.8}}+\sqrt{\frac{2\cdot 5.7}{9.8}}+\sqrt{\frac{2\cdot 4.5}{9.8}}$

$t=3.429\ s$

total time =3.429 s