Question

A ball is dropped from a height. If it takes $0.200s$s to cross the last $6.00m$ before hitting the ground, find the height from which it is dropped. Take $g=10m/s^2$.

Answer 1

Given:

The time taken by the ball to cover the last $6m$ is $0.2s$.

$s=6m,t=0.2s,$u=velocity just before 6m

The initial speed of the ball before falling the $6m$ will be given using second equation of motion:

$6=u\times 0.2+\frac{1}{2}\times g\times (0.2{)}^2$

$6=0.2u+5\times \frac{4}{100}$

$0.2u=6-0.2=5.8$

So, $u=29m/s$

If ball dropped from height, then the final speed

$v=\sqrt{2gH}$

$v=\sqrt{2g(H_m-6)}$

$29=\sqrt{2\times 10(H_m-6)}$

$H_m=48.05m$