From the top of a cliff 45 m above the sea, a boy throws a stone straight up at 12 m/s.

Step 1. Rise. $\dfrac{(12\ \text{m/s})^2}{2 \times 9.8\ \text{m/s}^2} = 7.35\ \text{m}$.

Step 2. Time. $4.9\ \text{m/s}^2 \times t^2 - 12\ \text{m/s} \times t - 45\ \text{m} = 0$ gives, with $\sqrt{144\ \text{m}^2/\text{s}^2 + 882\ \text{m}^2/\text{s}^2} = 32.0\ \text{m/s}$, $t = \dfrac{12\ \text{m/s} + 32.0\ \text{m/s}}{9.8\ \text{m/s}^2} = 4.49\ \text{s}$.

Step 3. Speed. $v^2 = (12\ \text{m/s})^2 + 2 \times 9.8\ \text{m/s}^2 \times 45\ \text{m} = 1026\ \text{m}^2/\text{s}^2$, so $v = 32.0\ \text{m/s}$, downwards.