A ball of mass $M$ falls from a height $h$ on a floor which the coefficient of restitution is $e$ . The height attained by the ball after two rebounds is:

e^{2} h

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e h^{2}

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e^{4} h

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\frac{h}{e^{4}}

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Solution

The correct option is C $e^{4} h$
A ball falling from height $h$ reaches the floors with velocity
$v = \sqrt{2 g h}$
By definition
$e = \frac{velocity of separation}{velocity of approach}$
$∴$ Rebound velocity $= e \sqrt{2 g h}$
$∴$ Rebound height $= e^{2} h$
After two rebounds, $h e i g h t = e^{4} h$ .

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A ball of mass M falls from a height h on a floor which the coefficient of restitution is e. The height attained by the ball after two rebounds is:

The correct option is C e4hA ball falling from height h reaches the floors with velocityv=√2ghBy definitione=velocity of separationvelocity of approach∴ Rebound velocity =e√2gh∴ Rebound height =e2hAfter two rebounds, height=e4h.

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