A small ball is projected from point A on the floor towards a vertical wall as shown in the figure. It hits the wall when its velocity is horizontal. Ball reaches point A after one bounce on the floor. If the coefficient of restitution is the same for the collision, find its value
 

Given BC is more height

When velocity becomes horizontal = ucosθ 

After collision;  v =eucosθ
and ${\mathrm{t}}_{\mathrm{BD}}=\frac{\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}(\because \mathrm{BC}\text{ is same })\therefore \mathrm{CD}=\frac{{\mathrm{eu}}^2\sin \mathrm{\theta cos}\mathrm{\theta }}{\mathrm{g}}$
Then; at D

Before collision        After collision $\left(\because {\mathrm{v}}_{\mathrm{y}}=\mathrm{e}.{\mathrm{u}}_{\mathrm{y}}\right)\therefore \mathrm{DA}=\frac{2{\mathrm{e}}^2{\mathrm{u}}^2\sin \mathrm{\theta cos}\mathrm{\theta }}{\mathrm{g}}$
From diagram: DA+DC=AC

$\begin{array}{l}\Rightarrow \frac{2{\mathrm{e}}^2{\mathrm{u}}^2\sin \mathrm{\theta cos}\mathrm{\theta }}{\mathrm{g}}+\frac{{\mathrm{eu}}^2\sin \mathrm{\theta cos}\mathrm{\theta }}{\mathrm{g}}=\frac{{\mathrm{u}}^2\sin \mathrm{\theta cos}\mathrm{\theta }}{\mathrm{g}}\\ \Rightarrow 2{\mathrm{e}}^2+\mathrm{e}-1=0\Rightarrow \mathrm{e}=\frac{1}{2}\end{array}$