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A tennis ball hits the floor with a speed $v$ at an angle $\theta$ with the normal to the floor. If the collision is inelastic and the co-efficient of restitution is $\varepsilon$, what will be the angle of reflection?
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$
Hint:
$\tan \theta^{\prime}=\frac{\operatorname{usin} \theta}{\operatorname{eucos} \theta} \quad \tan \theta^{\prime}=\frac{1}{\mathrm{e}} \tan \theta$
Since the floor is smooth. Hence tangential component of velocity remains unchanged.
$\tan \theta^{\prime}=\frac{\operatorname{usin} \theta}{\operatorname{eucos} \theta} \quad \tan \theta^{\prime}=\frac{1}{\mathrm{e}} \tan \theta$
Since the floor is smooth. Hence tangential component of velocity remains unchanged.
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