A ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it…
A ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it rebounds with a velocity $v_n$ and the ball rises to a height $h_n$, then coefficient of restitution is given by
$e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}$
$e=\left[\frac{h_0}{h_n}\right]^{1 / 2 n}$
$e=\frac{1}{n} \sqrt{\frac{h_n}{h_0}}$
$e=\frac{1}{n} \sqrt{\frac{h_0}{h_n}}$
Solution
The velocity of ball after $n$th rebound will be
$v_n=e^n v_0 \quad\left(\text { where } v_0=\sqrt{2 g h_0}\right)$
Therefore, the height after $n$th rebound will be
$\begin{aligned}
h_n & =\frac{v_n^2}{2 g}=\frac{e^{2 n} v_0^2}{2 g}=\frac{e^{2 n} \times 2 g h_0}{2 g} \\
\Rightarrow \quad e^{2 n} & =\frac{h_n}{h_0} \text { or } e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}
\end{aligned}$