A metal ball falls freely from a height h = 2 m onto a heavy…

A metal ball falls freely from a height $h = 2\ \mathrm{m}$ onto a heavy metal plate and bounces back without losing speed. Find the time between two successive impacts.

Explanation of Method

Time between bounces of a perfectly elastic ball.

Step 1: Symmetric bounce

It leaves the plate with the speed it arrived with, $\sqrt{2gh}$, so it rises back to $h$; rising and falling take equally long: $\Delta t = 2\sqrt{\frac{2h}{g}} = 2\sqrt{\frac{4}{9.81}}$.

\[\Delta t \approx 1.28\ \mathrm{s}\]

$y\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $2$ $\Delta t = 1.28\ \mathrm{s}$

Height against time. With no energy lost, each bounce is a mirror image of the fall: up for 0.64 s and down for 0.64 s.

Result

\[\boxed{\,\Delta t \approx 1.28\ \mathrm{s}\,}\]

Remark

Check. The original result is correct.