A ball can be thrown at 16 m/s over level ground…

Step 1. Greatest range. At $45^\circ$: $\dfrac{(16\ \text{m/s})^2}{9.8\ \text{m/s}^2} = 26.1\ \text{m}$.

Step 2. The two ranges. At $20^\circ$: $\dfrac{(16\ \text{m/s})^2 \times \sin 40^\circ}{9.8\ \text{m/s}^2} = 16.8\ \text{m}$. At $70^\circ$, $\sin 140^\circ = \sin 40^\circ$, so the range is also $16.8\ \text{m}$.

Step 3. The two times. $T_{20} = \dfrac{2 \times 16\ \text{m/s} \times \sin 20^\circ}{9.8\ \text{m/s}^2} = 1.12\ \text{s}$, and $T_{70} = \dfrac{2 \times 16\ \text{m/s} \times \sin 70^\circ}{9.8\ \text{m/s}^2} = 3.07\ \text{s}$.