Question

A ball is dropped from a roof of a building can reach at ground in 5 sec. If ball is stopped after 3 sec of its fall and then allowed to fall again, then find the time taken by the ball to reach ground for the remaining height.

Answer 1

Given,

$t_1=5s.t_2=3s$

So,

Let the total height be h,

$h=ut+\frac{1}{2}g{t}^2$

$=\frac{11}{2}g{t}_1^2=5\times 25=125m$ since $u=0$

$h^{\prime }=ut+\frac{1}{2}g{t}^2$

$\dfrac{1}{2}gt_2^2=5\times9=45m

Thus the momentum is stopped and the velocity is $0$

$h"=h-h^{\prime }=125-45$

$=80m$

since $h"=\frac{1}{2}g{t}^2$

$80=\frac{1}{2}\times 10t^2$

Thus, the time taken by the ball to reach ground for the remaining height. is $t=4s$