A ball rolls off the edge of a table 1.25 m high, moving horizontally at 3.0 m/s…

Step 1. Time. $t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 1.25}{9.8}} = 0.505\ \text{s}$.

Step 2. Distance. $x = 3.0\ \text{m/s} \times 0.505\ \text{s} = 1.52\ \text{m}$.

Step 3. Final velocity. $v_x = 3.0\ \text{m/s}$ and $v_y = 9.8\ \text{m/s}^2 \times 0.505\ \text{s} = 4.95\ \text{m/s}$ downwards. The speed is $\sqrt{(3.0\ \text{m/s})^2 + (4.95\ \text{m/s})^2} = 5.79\ \text{m/s}$, at $\tan^{-1}\dfrac{4.95\ \text{m/s}}{3.0\ \text{m/s}} = 58.8^\circ$ below the horizontal.