A particle of mass m, strikes the ground with angle of incidence ${45}^{°}$  with velocity V. if coefficient of restitution $e=\frac{1}{\sqrt{2}}$  the velocity of reflection is: 

After the reflection 

The component parallel to the surface will not alter but the component normal to the surface will be

${\mathrm{v}}_{\perp }=\mathrm{evcos\theta }$

Therefore, the velocity of reflection is

$v\text{'}=v\sqrt{{\sin }^2\theta +e^2{\cos }^2\theta }$

$\mathrm{v}\text{'}=\mathrm{v}\sqrt{\frac{1}{2}+{\left(\frac{1}{\sqrt{2}}\right)}^2\times \frac{1}{2}}=\frac{\sqrt{3}}{2}\mathrm{v}$