A glass ball collides with a smooth horizontal surface (xz plane) with a velocity$\mathrm{V}=\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}$. It the coefficient of restitution of collision be e, the velocity of the ball just after the collision will be

Collision takes place along the normal. Therefore the magnitude normal component (V_y) of the velocity of the glass ball is changed to ${\mathrm{V}}_{\mathrm{y}}^{\mathrm{\prime }}=-{\mathrm{eV}}_{\mathrm{y}}$ just after the collision where as the horizontal component (V_x) of its velocity remains constant due to the absence of any horizontal force.
$\Rightarrow$ The velocity of the ball just after the impact

$={\mathrm{V}}^{\mathrm{\prime }}={\overrightarrow{\mathrm{V}}}_{\mathrm{x}}^{\mathrm{\prime }}+{\overrightarrow{\mathrm{V}}}_{\mathrm{y}}^{\mathrm{\prime }}\Rightarrow {\overrightarrow{\mathrm{V}}}^{\mathrm{\prime }}={\mathrm{V}}_{\mathrm{x}}^{\mathrm{\prime }}\hat{\mathrm{i}}+{\mathrm{V}}_{\mathrm{y}}^{\mathrm{\prime }}\hat{\mathrm{j}}$

where  ${\mathrm{V}}_{\mathrm{x}}^{\mathrm{\prime }}=\mathrm{a}\text{ and }{\mathrm{V}}_{\mathrm{y}}^{\mathrm{\prime }}=\mathrm{eb}(\because \overrightarrow{\mathrm{V}}=\mathrm{a}\hat{\mathrm{i}}-\mathrm{b}\hat{\mathrm{j}})$

$\Rightarrow {\overrightarrow{\mathrm{V}}}^{\mathrm{\prime }}=\mathrm{a}\hat{\mathrm{i}}+\mathrm{eb}\hat{\mathrm{j}}$

Therefore the magnitude of the velocity ${\overrightarrow{\mathrm{V}}}^{\mathrm{\prime }}=\left|{\overrightarrow{\mathrm{V}}}^{\mathrm{\prime }}\right|=\sqrt{{\mathrm{a}}^2+{\mathrm{e}}^2{\mathrm{b}}^2}$

and the direction is given as $\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{V}}_{{\mathrm{x}}^{\mathrm{\prime }}}}{{\mathrm{V}}_{{\mathrm{y}}^{\mathrm{\prime }}}}\right)={\tan }^{-1}\left(\frac{\mathrm{a}}{\mathrm{eb}}\right)$ to the normal (vertical).