$A$ particle falling vertically from a height hits a plane surface inclined to the horizontal at an angle $\theta$ with speed $v_0$ and rebounds elastically. Find the distance along the plane where it will hit the second time.

(D) Let the point of impact be the origin $O$. We set the $x$-axis along the inclined plane (downwards) and the $y$-axis perpendicular to the inclined plane (upwards).
The initial velocity vector $\vec{v}_0$ is vertical downwards. Resolving this into components along the $x$ and $y$ axes:
$u_x = v_0 \sin \theta$
$u_y = -v_0 \cos \theta$
The acceleration due to gravity $g$ acts vertically downwards. Resolving this into components:
$a_x = g \sin \theta$
$a_y = -g \cos \theta$
For the particle to hit the plane again,the displacement along the $y$-axis must be zero $(y = 0)$.
Using $y = u_y t + \frac{1}{2} a_y t^2$:
$0 = (-v_0 \cos \theta) t + \frac{1}{2} (-g \cos \theta) t^2$
$0 = -t (v_0 \cos \theta + \frac{1}{2} g \cos \theta t)$
Since $t \neq 0$,we have $v_0 \cos \theta = -\frac{1}{2} g \cos \theta t$,which gives $t = \frac{2 v_0}{g}$.
Now,find the displacement along the $x$-axis at time $t = \frac{2 v_0}{g}$:
$x = u_x t + \frac{1}{2} a_x t^2$
$x = (v_0 \sin \theta) \left( \frac{2 v_0}{g} \right) + \frac{1}{2} (g \sin \theta) \left( \frac{2 v_0}{g} \right)^2$
$x = \frac{2 v_0^2 \sin \theta}{g} + \frac{1}{2} g \sin \theta \left( \frac{4 v_0^2}{g^2} \right)$
$x = \frac{2 v_0^2 \sin \theta}{g} + \frac{2 v_0^2 \sin \theta}{g} = \frac{4 v_0^2 \sin \theta}{g}$.