A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle $\alpha$ with the horizontal. Having fallen the distance $h$, the ball rebounds elastically off the inclined plane. At a distance $l=xh\sin \alpha$ from the impact point the ball rebounds for the second time. Find $x$.

The ball strikes the inclined plane $\left(Ox\right)$ at point $O$ (origin) with velocity $v_0=\sqrt{2gh}$ ........ (1)
As the ball elastically rebounds, it recalls with same velocity $v_0$, at the same angle $\alpha$ from the normal or y axis (figure shown below). Let the ball strikes the incline second time at $P$, which is at a distance $l$ (say) from the point $O$, along the incline. From the equation
$y=v_{0y}t+\frac{1}{2}{w}_y{t}^2$
$0=v_0\cos \alpha \tau -\frac{1}{2}g\cos \alpha {\tau }^2$
where $\tau$ is the time of motion of ball in air while moving from $O$ to $P$.
As $\tau \ne 0$, so, $\tau =\frac{2v_0}{g}$................. (2)
Now from the equation.
$x=v_{0x}t+\frac{1}{2}{w}_x{t}^2$
$l=v_0\sin \alpha \tau +\frac{1}{2}g\sin \alpha {\tau }^2$
so, $l=v_0\sin \alpha \left(\frac{2v_0}{g}\right)+\frac{1}{2}g\sin \alpha {\left(\frac{2v_0}{g}\right)}^2$
$=\frac{4v_0^2\sin \alpha }{g}$ (using 2)
Hence, the sought distance, $l=\frac{4\left(2gh\right)\sin \alpha }{g}=8h\sin \alpha$ (Using equation 1)