A massive vertical wall is approaching a man at a speed u. When it is at a distance of 10 m, the man throws a ball with speed 10 m/s at an angle of 37°, which after having a completely elastic collision with the wall, reaches back directly into the hands of the man. The velocity of the wall is

As the vertical motion remains unaffected
$\begin{array}{l}\mathrm{T}=\frac{2\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}\Rightarrow \mathrm{T}=\frac{2\times 10\times 3/5}{10}\\ \mathrm{T}=\frac{6}{5}\mathrm{S} ...\left(1\right)\end{array}$
From coefficient of restitution along the line of collision
$\text{c=}\frac{\text{relative\hspace{0.17em}velocity\hspace{0.17em}of\hspace{0.17em}separation\hspace{0.17em}after\hspace{0.17em}collision}}{\text{relative\hspace{0.17em}velocity\hspace{0.17em}of\hspace{0.17em}apporochbefore\hspace{0.17em}collision}}$
$\text{1=}\frac{\mathrm{v}-\mathrm{u}}{8+\mathrm{u}}$
$\mathrm{v}-2\mathrm{u}=8...\left(2\right)$
By the time $\left({\mathrm{t}}_1\right)$ ball collide will wall, distance travelled by man= $10-{\mathrm{ut}}_1$
${\mathrm{t}}_1=\frac{10}{8+\mathrm{u}}.....\left(3\right)$
By the time $\left({\mathrm{t}}_2\right)$ ball reach the hand of man  ${\mathrm{t}}_2=\frac{10-{\mathrm{u}}_1}{\mathrm{v}}.....\left(4\right)$
Since $\mathrm{T}={\mathrm{t}}_1+{\mathrm{t}}_{2....\left(5\right)}$
From equation  $\left(2\right),\left(3\right),\left(4\right),\left(5\right)$
$\mathrm{u}=\frac{13}{3}\mathrm{m}/\mathrm{s}$