A body falls from a height '$h$' on a horizontal surface and rebounds. Then it falls again and rebounds and so on. If the coefficient of restitution is $\frac{1}{2}$, then the total distance travelled by the ball before coming to rest is (neglect air resistance)

The correct option is D $\frac{5h}{3}$
When a body falls from height $h$ it hits the ground with the velocity $v=\sqrt{2gh}$

$\Rightarrow h=\frac{v^2}{2g}$

Now the ball undergoes collision, loses some energy and rebounds with a velocity $ev$ where $e$ is the coefficient of restitution. It will now rise to a height $h_1$ and come back by covering the same distance $h_1$

$h_1=\frac{(ev{)}^2}{2g}$

Now the ball will again strike the ground and undergo collision and will rebound with a velocity $e^{2}v$. If it rises to a height $h_2$

$h_2=\frac{(e^{2}v{)}^2}{2g}$

Thus, the total distance is given as

$d=h+2h_1+2h_2+...$

$d=\frac{v^2}{2g}+\frac{2e^2{v}^2}{2g}+\frac{2e^4{v}^2}{2g}+...$

$\Rightarrow d=\frac{v^2}{2g}+\frac{2e^2{v}^2}{2g}(1+e^2+e^4+...)$

Here $(1+e^2+e^4+...)$ forms an infinite GP with first term $a=1$ and common rartio $r=e^2$

So, sum $S=\frac{a}{1-r}=\frac{1}{1-e^2}$

$\Rightarrow d=\frac{v^2}{2g}+\frac{2e^2{v}^2}{2g}\frac{1}{1-e^2}$

$\Rightarrow d=\frac{v^2}{2g}[1+\frac{2e^2}{1-e^2}]$

$\Rightarrow d=\frac{v^2}{2g}\left[\frac{1-e^2+2e^2}{1-e^2}\right]$

$d=h\left[\frac{1+e^2}{1-e^2}\right]$

$d=h\left[\frac{1+{\left(\frac{1}{2}\right)}^2}{1-{\left(\frac{1}{2}\right)}^2}\right]$ ($\because e=\frac{1}{2}$)

$d=h\left[\frac{5/4}{3/4}\right]$

$d=\frac{5h}{3}$