A ball of mass m hits the floor with a speed v making an angle of...

The parallel component of velocity of the ball remains unchanged. This gives $v^{\prime} \sin \theta^{\prime}=v \sin \theta$ ...(i) For the components normal to the floor, the velocity of separation $=v^{\prime} \cos \theta^{\prime}$ and the velocity of approach $=v \cos \theta$ Hence, $\quad v^{\prime} \cos \theta^{\prime}=e v \cos \theta$ ... (ii) From Eqs. (i) and (ii) $$\begin{aligned} v^{\prime} & =v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta} \\ \text{ and } \quad \tan \theta^{\prime} & =\frac{\tan \theta}{e}\end{aligned}$$ For elastic collision, $e=1$, so that $\theta^{\prime}=\theta$ and $v^{\prime}=v$