A ball of mass $m$ strikes a rigid wall with speed $u$ at an angle of ${30}^{\circ }$ and get reflected with the same speed and at the same angle as shown in the figure. If the ball is in contact with the wall for time t, then the force acting on the wall is

Initial momentum of the ball is ${\overrightarrow{p}}_i=mu\cos {30}^{\circ }\hat{i}-mu\sin {30}^{\circ }\hat{j}$

Final momentum of the ball is ${\overrightarrow{\mathrm{p}}}_{\mathrm{f}}=-\mathrm{mucos}{30}^{\circ } \hat{\mathrm{i}}-\mathrm{musin}{30}^0 \hat{\mathrm{j}}$

Therefore change in momentum, $∆\overrightarrow{p}={\overrightarrow{p}}_f-{\overrightarrow{p}}_i=-2mu\cos {30}^{\circ }\hat{i}$

Impulse $=$Change in momentum $=-2mu\cos {30}^{\circ }\hat{i}$

As impulse and force are in the same direction, therefore, force on the ball due to the wall is normal to the wall along the negative $x-$axis. Using Newton's $3^{rd}$ law of motion the force on the wall due to the ball is normal to the wall along the positive $x-$direction. 

Therefore $F=\frac{2mu\cos {30}^{\circ }}{t}$