Question

A body is thrown vertically upwards from ground with a speed of $10m/s$. If coefficient of restitution of ground is, $e=1/2$ and the time elapsed (after the ball has been thrown) when it is at its subsequent maximum height for the third time is $\frac{x}{4}sec$, find the value of $x$.

Answer 1

The maximum height the ball will reach is $V^2=U^2+2aS$

$a=-10m/s^2$ deceleration

$0=100-2\times 10\times S$

$S=5m$

$v=u+at$

$0=10-10t$

$t=1sec$ for the ball to reach the top and same time to reach to ground

After collision ball will attain the velocity of

$e=\frac{V_{after}}{V_{before}}$

$V_{after}=\frac{V_{before}}{2}=\frac{10}{2}=5m/s$

therefore time taken to reach $h^{{}^{\prime }}$ is

$v=u+at$

$0=5-10t$

$t=0.5s$

now ball will again hit the ground and attain new maximum height

$e=\frac{v_{after}}{5}$

$v_{after}=2.5m/s$

time taken after second collision to reach maximum height is

$v=u+at$

$0=2.5-10t$

$t=0.25s$

therefore total time taken is

$T=t_1\times 2+t_2\times 2+t_3$

$t_1$ and $t_2$ are multiplied by 2 because it takes same time for ball to attain maximum height and come back to ground

$T=1\times 2+0.5\times 2+0.25=3.25s=\frac{325}{100}=\frac{13}{4}s$