A ball is dropped from a height $h$ above a point on an inclined plane with an angle of inclination $\theta$. The ball makes an elastic collision with the surface and rebounds off the plane. Determine the distance from the point of first impact to the point where ball hits the plane second time.

The correct option is D $8h\sin \theta$
The situation is shown in figure below. We take the point of first impact as the origin of our referance.

It is given that the ball rebounds elastically ,it implies that there is no change in kinetic energy of the ball before and after the collision.

The ball rebounds with the same velocity with which it will strike the plane after falling a distance $h$, which is
$u=\sqrt{2gh}$.


After rebound, the particle will be accelerated by $g\sin \theta$ and will be retarded by $g\cos \theta$.

Here time of flight from first impact to the second impact is given as

$T_f=\frac{2u_y}{a_y}=\frac{2u\cos \theta }{g\cos \theta }=\frac{2u}{g}$

In this duration the distance travelled by the particle is

$R=u_x{T}_f+\frac{1}{2}{a}_x{T}_f^2$

$\Rightarrow R=\left(u\sin \theta \right)\times \frac{2u}{g}+\frac{1}{2}\left(g\sin \theta \right){\left(\frac{2u}{g}\right)}^2$

$\Rightarrow R=\frac{4u^2\sin \theta }{g}$

As, $u=\sqrt{2gh}$

$\Rightarrow R=\frac{4(\sqrt{2gh}{)}^2\sin \theta }{g}$

$\therefore R=8h\sin \theta$

Hence, option (d) is the correct answer.