A ball falls freely from rest on to a hard horizontal floor and repeatedly bounces. If the velocity of the ball just before the first bounce is $7 \mathrm{~ms}^{-1}$ and the coefficient of restitution is 0.75 . The total distance travelled by the ball before it comes to rest is (acceleration due…

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A ball falls freely from rest on to a hard horizontal floor and repeatedly bounces. If the velocity of the ball just before the first bounce is $7 \mathrm{~ms}^{-1}$ and the coefficient of restitution is 0.75 . The total distance travelled by the ball before it comes to rest is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Aicharya · Accepted Answer

$\mathrm{l}=0.75, \mathrm{u}=7 \mathrm{~ms}^{-1}$ Total distance traveled, $\mathrm{S}=\mathrm{H}+2 \mathrm{H}_1+2 \mathrm{H}_2+\ldots \ldots \ldots$ $\begin{aligned} & =\frac{u^2}{2 g}+2 \cdot \frac{(e u)^2}{2 g}+2 \cdot \frac{\left(e^2 u\right)^2}{2 g}+\ldots .+\infty \\ & =\frac{u^2}{2 g}\left(1+2 e^2+2 e^4+\ldots . .+\infty\right) \\ & =\frac{u^2}{2 g}+\frac{e^2 u^2}{g}\left(1+e^2+e^4+\ldots+\infty\right) \\ & =\frac{u^2}{2 g}+\frac{e^2 u^2}{g}\left(\frac{1}{1-e^2}\right)=\frac{u^2}{g}\left(\frac{1}{2}+\frac{e^2}{1-e^2}\right) \\ & =\frac{u^2}{2 g}\left(\frac{1+e^2}{1-e^2}\right)=\frac{7 \times 7}{2 \times 10}\left[\frac{1+(0.75)^2}{1-(0.75)^2}\right]\end{aligned}$

A ball falls freely from rest on to a hard horizontal floor and repeatedly bounces. If the velocity of the…

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