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Question

The ball has an initial velocity as $\sqrt{2 g h}$

Thus the velocity just before it collides on the ground is given using $v^{2}=u^{2}+2 a s$

or

Vedclass · Accepted Answer

$1 / \sqrt 2$

Vedclass · Answer

The ball has an initial velocity as $\sqrt{2 g h}$

Thus the velocity just before it collides on the ground is given using $v^{2}=u^{2}+2 a s$

or

$v^{2}=2 g h+2 g h=4 g h$

or

$v_{i}=\sqrt{4 g h}$

Now the velocity changes to $v_{f}$ (say) after collision.

The ball just reaches the initial height $h.$

Thus we get

$0=v_{f}^{2}-2 g h$

or $v_{f}=\sqrt{2 g h}$

Thus the coefficient of restitution is given as

$e=\frac{v_{f}}{v_{i}}=\frac{\sqrt{2 g h}}{\sqrt{4 g h}}=\frac{1}{\sqrt{2}}$

$A$ ball is thrown vertically downwards with velocity $\sqrt {2gh} $ from $a$ height $h$. After colliding with the ground it just reaches the starting point. Coefficient of restitution is

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