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A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle $α$ with the horizontal. Having fallen the distance h, the ball rebounds on the inclined plane. The distance along the plane from impact point to which ball rebound for the second time is $x h \sin α$ (assuming collision is elastic) The value of $x$ is

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answer is 8.

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Detailed Solution

The velocity of the ball just before hitting the plane is $u = \sqrt{2 g h}$
Using equation of motion along plane and perpendicular to plane $y = u_{y} t - \frac{1}{2} a_{y} t^{2}$

$∴ x = \frac{4 {(\sqrt{2 g h})}^{2} \sin α}{g} = 8 h \sin α$

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