A ball is thrown at 10 m/s, at 30° above the horizontal, from the edge of a flat roof 20 m high…

Step 1. Split. $u_x = 10\ \text{m/s}\cos 30^\circ = 8.66\ \text{m/s}$ and $u_y = 10\ \text{m/s}\sin 30^\circ = 5\ \text{m/s}$.

Step 2. The equation. $-20\ \text{m} = 5\ \text{m/s} \times t - 4.9\ \text{m/s}^2 \times t^2$, so $4.9\ \text{m/s}^2 \times t^2 - 5\ \text{m/s} \times t - 20\ \text{m} = 0$.

Step 3. Solve. Here $(5\ \text{m/s})^2 + 4 \times 4.9\ \text{m/s}^2 \times 20\ \text{m} = 417\ \text{m}^2/\text{s}^2$, whose square root is $20.4\ \text{m/s}$, so $t = \dfrac{5\ \text{m/s} + 20.4\ \text{m/s}}{9.8\ \text{m/s}^2} = 2.59\ \text{s}$. The other root is negative: a time before the throw.

Step 4. Distance. $x = 8.66\ \text{m/s} \times 2.59\ \text{s} = 22.5\ \text{m}$.