A Bouncy Problem 3

When the ball falls from rest at a height $h$ , its velocity upon hitting for the first time is given by $v_0^2 = 0^2 + 2gh$ , where $g$ the acceleration due to gravity. Therefore, $v_0 = \sqrt{2gh}$ . If $e$ is the coefficient of restitution, then the first rebound velocity after hitting the ground $v_1 = ev_2 = e\sqrt{2gh}$ , second rebound velocity, $v_2 = e^2\sqrt{2gh}$ and third rebound velocity, $v_3 = e^3\sqrt{2gh}$ . After rebounding the third time, the maximum height reached is $\dfrac h2$ . Therefore,

$\begin{aligned} 0^2 & = v_3^2 - 2g\left(\frac h2 \right) \\ v_3^2 & = 2g\left(\frac h2 \right) \\ e^6 (2gh) & = gh \\ \implies e & = \frac 1{\sqrt[6]2} \approx \boxed{0.86} \end{aligned}$