A ball thrown over level ground rises to a greatest height of 12 m and lands 30 m away…

Step 1. Angle. $\tan\theta = \dfrac{4 \times 12\ \text{m}}{30\ \text{m}} = 1.6$, so $\theta = 58.0^\circ$.

Step 2. The vertical part. $u\sin\theta = \sqrt{2 \times 9.8\ \text{m/s}^2 \times 12\ \text{m}} = 15.3\ \text{m/s}$.

Step 3. Speed. $u = \dfrac{15.3\ \text{m/s}}{\sin 58.0^\circ} = 18.1\ \text{m/s}$.