Hitting a Target

The question

A ball kicked at 16 m/s must land 20 m away. Which angles work?

Two angles: 25° and 65°. The range $R={\displaystyle \frac{v^2\sin 2\alpha }{g}}$ equals 20 m when $\sin 2\alpha =0.766$, which happens for $2\alpha =50\mathrm{°}$ and for $2\alpha =130\mathrm{°}$. The flat shot at 25° arrives after 1.38 s; the lob at 65° takes 2.96 s. The further the target, the closer the two angles get, until they meet at 45°.

How to find the angles for a given range

1. Write the range as $R={\displaystyle \frac{v^2\sin 2\alpha }{g}}$.
2. Set it equal to the distance and solve for $\sin 2\alpha$.
3. Take both solutions: $2\alpha$ and ${180}^{\circ }-2\alpha$.
4. Halve them to get the two angles. If $\sin 2\alpha$ comes out bigger than 1, the target is out of reach.