A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find the direction of velocity (angle with horizontal) with which it strikes the ground.

$$\begin{gathered} v_x=u_x=20 \mathrm{m}/\mathrm{s} \\ \text{Time taken by the ball to reach the ground }t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 100}{10}}=\sqrt{20} \mathrm{s} \\ {v}_y=u_y+a_{y}t = 0-10\sqrt{20} = -10\sqrt{20} \mathrm{m}/\mathrm{s} \\ \theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)=-{\tan }^{-1}\left(\sqrt{5}\right) \\ \text{∴ Direction of velocity is }{\tan }^{-1}\left(\sqrt{5}\right)\text{ below horizontal.} \end{gathered}$$