A fielder throws a cricket ball straight up at 18 m/s to celebrate a catch…

Step 1. To the top. $t = \dfrac{18\ \text{m/s}}{9.8\ \text{m/s}^2} = 1.84\ \text{s}$.

Step 2. Height. $H = \dfrac{(18\ \text{m/s})^2}{2 \times 9.8\ \text{m/s}^2} = 16.5\ \text{m}$.

Step 3. Back. $2 \times 1.84\ \text{s} = 3.67\ \text{s}$.