A stone is thrown horizontally at 15 m/s from the top of a cliff 45 m high, out over the sea…

Step 1. Time. $t = \sqrt{\dfrac{2 \times 45\ \text{m}}{9.8\ \text{m/s}^2}} = 3.03\ \text{s}$.

Step 2. Distance. $x = 15\ \text{m/s} \times 3.03\ \text{s} = 45.5\ \text{m}$.

Step 3. Landing velocity. $v_x = 15\ \text{m/s}$ and $v_y = 9.8\ \text{m/s}^2 \times 3.03\ \text{s} = 29.7\ \text{m/s}$ downwards. The speed is $\sqrt{(15\ \text{m/s})^2 + (29.7\ \text{m/s})^2} = 33.3\ \text{m/s}$, at $\tan^{-1}\dfrac{29.7\ \text{m/s}}{15\ \text{m/s}} = 63.2^\circ$ below the horizontal.