A ball is projected from the ground with speed $u$ at an angle $\alpha$ with horizontal. It collides with a wall at distance $a$ from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall.

The correct options are
B $\frac{1}{(\frac{u^2\sin 2\alpha }{ag}-1)}$

F $\frac{1}{(\frac{u^2\sin 2\alpha }{ag}-1)}$

We know that the horizontal component of the velocity of ball during the path $\text{OAB}$ is $u\cos \alpha$.
While in its return journey $\text{BCO}$ it is $eu\cos \alpha$. The time of flight $T$ also remains unchanged.


Hence, time of flight,
$T=t_{OAB}+t_{BCO}.....\left(1\right)$

where,
$t_{OAB}=$ Time taken to reach the point $\text{B}$ along horizontal direction $=\frac{a}{ucos\alpha }$.

$t_{BCO}=$Time taken to return point $\text{O}$ along horizontal direction $=\frac{a}{eu\cos \alpha }$.

Substituting the values in equation $\left(1\right)$, for the given projectile motion, we get

$\frac{2u\sin \alpha }{g}=\frac{a}{u\cos \alpha }+\frac{a}{eu\cos \alpha }$

$\Rightarrow \frac{a}{eu\cos \alpha }=\frac{2u\sin \alpha }{g}-\frac{a}{u\cos \alpha }$

$\Rightarrow \frac{a}{eu\cos \alpha }=\frac{2u^2\sin \alpha \cos \alpha -ag}{gu\cos \alpha }$

$\Rightarrow e=\frac{ag}{2u^2\sin \alpha \cos \alpha -ag}$

$\Rightarrow e=\frac{1}{(\frac{u^2\sin 2\alpha }{ag}-1)}$

Option (b) is correct answer.

Why this question ? This question checks the concept of oblique collision.