A ball collides with a fixed inclined plane of inclination θ after falling through a distance h . If it moves horizontally just after the impact the coefficient of restitution is

Writing coefficient of restitution equation
$$\begin{gathered} \mathrm{e}=\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{u}}_1-{\mathrm{u}}_2} \\ \mathrm{initial} \mathrm{velocity} \mathrm{be} {\mathrm{v}}_0; \end{gathered}$$

resolve into components, as ${\mathrm{v}}_0\mathrm{cos\theta }$ perpendicular to the plane downward $={\mathrm{u}}_1$ and   ${\mathrm{v}}_0\mathrm{sin\theta }$ parallel to the plane downward

final velocity be v ; resolving into components, as $\mathrm{vcos\theta }$ parallel to the plane and $\mathrm{vsin\theta }$ perpendicular to the plane upward $={\mathrm{v}}_1$

on substituting 

$$\begin{gathered} \mathrm{e}=\frac{0-\mathrm{vsin\theta }}{-{\mathrm{v}}_0\mathrm{cos\theta }-0} \\ \Rightarrow \mathrm{e}=\frac{\mathrm{v}}{{\mathrm{v}}_0}\mathrm{tan\theta }---\left(1\right) \end{gathered}$$

The velocity of ball will remain unchanged in the direction of sloping surface (common tangent)
$V_0 \sin \theta = v \cos \theta$
$\begin{array}{l}\Rightarrow \frac{\mathrm{v}}{{\mathrm{v}}_0}=\mathrm{tan\theta }---\left(2\right)\\ \mathrm{substitute} \mathrm{eqn}\left(2\right) \mathrm{in} \mathrm{eqn}\left(1\right)\\ \Rightarrow \mathrm{e}={\tan }^2\mathrm{\theta }\end{array}$