A ball is given velocity $\mathrm{u}=\sqrt{3\mathrm{gd}}$ at an angle 45^o with horizontal. It strikes a wall at distance d and returns to its original position. Find the coefficient of resistance e

The velocity remains same in the vertical direction (along the surface) and the velocity become e times and opposite in the horizontal direction (perpendicular to the surface)
 

Time from O to A, $t_1=\frac{d}{u\cos {45}^{\circ }}$
Time from A to O, ${\mathrm{t}}_2=\frac{\mathrm{d}}{\mathrm{eucos}{45}^{\circ }}$
Total time $\mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2=\frac{\mathrm{d}}{\mathrm{eucos}{45}^{\circ }}\left(1+\frac{1}{\mathrm{e}}\right)$
Time for vertical motion (O to O)
$\begin{array}{l}0=\mathrm{usin}{45}^{\circ }\mathrm{T}-\frac{1}{2}{\mathrm{gT}}^2\Rightarrow \mathrm{T}=\frac{2\mathrm{usin}{45}^{\circ }}{\mathrm{g}}\\ \mathrm{t}=\mathrm{T}\\ \frac{\mathrm{d}}{\mathrm{ucos}{45}^{\circ }}\left(1+\frac{1}{\mathrm{e}}\right)=\frac{2\mathrm{usin}{45}^{\circ }}{\mathrm{g}}\\ \mathrm{d}\left(1+\frac{1}{\mathrm{e}}\right)=\frac{2{\mathrm{u}}^2\sin {45}^{\circ }\cos {45}^{\circ }}{\mathrm{g}}\\ =\frac{{\mathrm{u}}^2}{\mathrm{g}}\sin {90}^{\circ }=\frac{3\mathrm{gd}}{\mathrm{g}}=3\mathrm{d}\\ 1+\frac{1}{\mathrm{e}}=3\\ \mathrm{e}=\frac{1}{2}\end{array}$