<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
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{\bf Question}

A rubber ball is dropped from rest onto a tile floor a distance
$h_1$ away.  The ball bounces up to a height $h_2$.  What is the
coefficient of restitution?

\vspace{.25in}

{\bf Answer}

The ball falls under gravity:

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By Newton's 2nd law:
\begin{eqnarray*} m \ddot y &amp; = &amp; -mg \\ mv \frac{dv}{dy} &amp; =
&amp; -mg \hspace {.1in} (v = \dot y) \\ \frac{1}{2} v^2 &amp; = &amp; -gy +
gh_1 \end{eqnarray*}

Hence on impact with the floor $\ds v = v_0 = \sqrt{2gh_1} $

Also if the ball has speed $v_1$ (upwards) just after impact, $\ds
v_1 = \sqrt{2gh_2}$

Now by differentiation  $\ds v_1 = ev_0 \Rightarrow e =
\frac{v_1}{v_0} = \sqrt{\frac{h_2}{h_1}}$



\end{document}
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