A ball is projected with velocity ${\text{v}}_0=\sqrt{\text{gh}}$ at an angle$\text{'}\theta \text{'}$ to the horizontal towards a vertical smooth wall distant ‘d’ from the point of projection. After collision the ball returns to the point of projection. The coefficient of restitution is

${\text{t}}_{\text{AB}}=\frac{\text{d}}{{\text{v}}_{\text{0}}\cos \theta }$

We know that ${\text{t}}_{\text{AB}}+{\text{t}}_{\text{BA}}=\frac{2{\text{v}}_0\sin \theta }{\text{g}}$

After collision between ball and wall

${\text{v}}_x=\text{e}{\text{.u}}_x={\text{ev}}_{\text{0}}\cos \theta$

Distance$\text{d}=\text{eu}\cos \theta {\text{t}}_{\text{BA}}$

$\left({\text{s}}_x={\text{u}}_x\text{t}\right)$

${\text{t}}_{\text{BA}}=\frac{\text{d}}{{\text{ev}}_{\text{0}}\cos \theta }$

$\frac{\text{d}}{{\text{v}}_{\text{0}}\cos \theta }\left(1+\frac{1}{\text{e}}\right)=\frac{{\text{v}}_{\text{0}}^{\text{2}}\sin \theta }{\text{g}}$

$\text{d}\left(1+\frac{1}{\text{e}}\right)=\frac{{\text{v}}_{\text{0}}^{\text{2}}\sin \theta }{\text{g}}$

Given${\text{v}}_{\text{0}}=\sqrt{\text{gh}}$

$\text{d}\left(1+\frac{1}{\text{e}}\right)=\text{h}\sin 2\theta$

$1+\frac{1}{\text{e}}=\frac{\text{h}}{\text{d}}\sin 2\theta$

By solving$\text{e}=\frac{\text{d}}{\text{h}\sin 2\theta -\text{d}}$