A body falls from a height h on a horizontal surface and rebounds. Then it falls again and again rebounds and so on. If the restitution coefficients is $\frac{1}{3}$ , the total distance covered by the body before it comes to rest is

\frac{h}{4}

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\frac{5 h}{4}

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2 h

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3 h

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Solution

The correct option is A $\frac{5 h}{4}$
$Explanation:-$

Total Distance $= h + 2 h_{1} + 2 h_{2} + 2 h_{3} + . . . .$

$= h + \frac{2 e^{2} v^{2}}{2 g} + \frac{2 e^{4} v^{2}}{2 g} + . . . . . .$

$= \frac{v^{2}}{2 g} + \frac{2 e^{2} v^{2}}{2 g} [1 + e^{2} + e^{4} + . . . .]$

$= \frac{v^{2}}{2 g} + \frac{2 e^{2} v^{2}}{2 g} [\frac{1}{1 - e^{2}}]$

$= \frac{v^{2}}{2 g} [\frac{1 - e^{2} + 2 e^{2}}{1 - e^{2}}]$

$= \frac{v^{2}}{2 g} [\frac{1 + e^{2}}{1 - e^{2}}]$

$= h ⎡ ⎢ ⎢ ⎢ ⎣ \frac{1 + \frac{1}{9}}{1 - \frac{1}{9}} ⎤ ⎥ ⎥ ⎥ ⎦$

$= h \times ⎡ ⎢ ⎢ ⎢ ⎣ \frac{\frac{10}{9}}{\frac{8}{9}} ⎤ ⎥ ⎥ ⎥ ⎦$

$= \frac{5 h}{4}$

$Hence the correct option is B$

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