A ball is projected in a direction inclined to the vertical and bounces on a smooth horizontal plane. The range of one rebound is R. If the coefficient of restitution is e, then range of the next rebound is :

R^{'} = e R

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R^{'} = e^{2} R

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R^{'} = \frac{R}{e}

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R^{'} = R

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Solution

The correct option is A $R^{'} = e R$
$T i m e o f f l i g h t f o r f i r s t r e b o u n d, T_{1} = \frac{2 u s i n θ}{g} T i m e o f f l i g h t f o r s e c o n d r e b o u n d, T_{2} = \frac{2 u e s i n θ}{g} b e c u a s e o f t h e c o e f f i c i e n t o f r e s t i t u t i o n v e r t i c a l v e l o c i t y b e c o m e s u e s i n θ . R a n g e o f f i r s t r e b o u n d, R_{1} = v c o s θ \times T_{1} R a n g e o f s e c o n d r e b o u n d, R_{2} = v c o s θ \times T_{2} = v c o s θ \frac{2 u e s i n θ}{g} \Rightarrow R_{2} = e R_{1}$

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