A ball is dropped on to a horizontal plate from a height $h=9 \mathrm{m}$ above it. If the coefficient of restitution is $e=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, the total distance travelled before the ball comes to rest is

The impact between the ball and the plate is direct impact. If u is the velocity of the ball before impact, its velocity after is eu, where e is the coefficient of restitution.

But $u=\sqrt{2\text{gh}}$ since the ball falls from a height h. So the velocity of the ball at the first rebound

$={\text{v}}_1=eu=e\sqrt{2\text{gh}}$

The height h₁ to which it rises after first rebound

$h_1=\frac{{\text{v}}_1^2}{2\text{g}}=\frac{e^2·2\text{gh}}{2\text{g}}=e^{2}h$

The velocity v₂ after second rebound v₂ = ev₁ = e²u

The height h₂ it rises after second rebound

$h_2=\frac{{\text{v}}_2^2}{2\text{g}}=\frac{e^4·u^2}{2\text{g}}=\frac{e^4·2\text{gh}}{2\text{g}}=e^{4}h$

In general, the height hn to which the ball rises after n^th rebound is given by h_n = e²ⁿh.

Total distance travelled

    = h + 2h₁ + 2h₂ + 2h₃ + ................+2h_n+............to ∞

    = h + 2 (e²h + e⁴h + e⁶ h +.......)

    = h + 2he²(1 + e² + e⁴+...........)

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

    $=\frac{9\left(1+{\left(\frac{1}{2}\right)}^2\right)}{\left(1-{\left(\frac{1}{2}\right)}^2\right)}=15\text{m}$