A ball is dropped from height $H$ on to a horizontal surface. If the coefficient of restitution is $e$ , then the total time after which it comes to rest is

\sqrt{\frac{2 H}{g}} (\frac{1 - e}{1 + e})

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\sqrt{\frac{2 H}{g}} (\frac{1 + e}{1 - e})

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\sqrt{\frac{2 H}{g}} (\frac{1 + e^{2}}{1 - e^{2}})

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\sqrt{\frac{2 H}{g}} (\frac{1 - e^{2}}{1 + e^{2}})

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Solution

The correct option is B $\sqrt{\frac{2 H}{g}} (\frac{1 + e}{1 - e})$
If a body drops from height $h_{1}$ and bounces back to $h_{2}$ then

$e = \sqrt{\frac{h_{2}}{h_{1}}}$

$h_{2} = e^{2} h_{1}$
time elapsed between falling from height $h_{1}$ and going back to $h_{2}$

$\frac{2 \times \sqrt{2 g h_{2}}}{g} = t_{1}$

$t_{2} = \frac{2 \times \sqrt{2 g h_{3}}}{g}$

$\sum_{n = 1}^{n = \infty} t_{i} = \frac{2}{g} \times \sqrt{2 g} (\sqrt{h_{n}} + \sqrt{h_{n - 1}} . . . . . . . . \sqrt{h_{2}}) + \frac{\sqrt{2 g H}}{g}$

$= \frac{2}{g} \times \sqrt{2 g} (e^{n - 1} . . . . . . . . . . . e^{2} \sqrt{h_{1}} + e \sqrt{h_{1}}) + \frac{\sqrt{2 g H}}{g}$

$= \frac{2}{g} \times \sqrt{2 g} \times \sqrt{h_{1}} \times \frac{1}{1 - e} + \frac{\sqrt{2 g H}}{g}$

now $\sqrt{h_{1}} = e \sqrt{H}$

$t \Rightarrow t = \frac{2}{g} \sqrt{2 g} \times e \sqrt{H} \times \frac{1}{1 - e} + \frac{\sqrt{2 g H}}{g} = \frac{\sqrt{2 g H}}{g} (\frac{1 + e}{1 - e})$

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