A girl throws a ball with initial velocity v at an  inclination of 45°. The ball strikes the smooth vertical wall at a horizontal distance d from girl and after rebounding returns to her hand. What is the coefficient of restitution between wall and ball ?

See fig.

As the wall is smooth, the vertical component of the impulse it receives is zero, The total time of flight is given by

$0=\left(\mathrm{vsin}{45}^{\circ }\right)\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2\text{ or }\mathrm{t}=\frac{2\mathrm{vsin}{45}^{\circ }}{\mathrm{g}}$

The time of flight t₂ from G to A

$=\frac{\mathrm{d}}{\mathrm{vcos}{45}^{\circ }}$

The time of flight t₂ from A to G

$=\frac{\mathrm{d}}{\mathrm{evcos}{45}^{\circ }}$

where e = coefficient of restitution.

$\text{ So, }\frac{2v \sin {45}^{\circ }}{\mathrm{g}}=\frac{\mathrm{d}}{\mathrm{Vcos}{45}^{\circ }}+\frac{\mathrm{d}}{\mathrm{evcos}{45}^{\circ }}$

$\begin{array}{rl}\mathrm{or} & \frac{\sqrt{2}\mathrm{v}}{\mathrm{g}}=\frac{\mathrm{d}\sqrt{2}}{\mathrm{v}}+\frac{\mathrm{d}\sqrt{2}}{\mathrm{ev}}\\ & \frac{\mathrm{v}}{\mathrm{g}}=\frac{\mathrm{d}}{\mathrm{V}}(1+\frac{1}{\mathrm{e}})\text{or}\frac{{\mathrm{v}}^2}{\mathrm{gd}}=1+\frac{1}{\mathrm{e}}\end{array}$

$\text{ Solving, we get }\mathrm{e}=\left(\frac{\mathrm{gd}}{{\mathrm{v}}^2-\mathrm{gd}}\right).$