A ball is thrown straight up at 16 m/s. Ignoring the air: At what height above the hand is its…

Step 1. Height. $(8\ \text{m/s})^2 = (16\ \text{m/s})^2 - 2 \times 9.8\ \text{m/s}^2 \times y$, so $y = \dfrac{192\ \text{m}^2/\text{s}^2}{19.6\ \text{m/s}^2} = 9.80\ \text{m}$.

Step 2. Going up. $8\ \text{m/s} = 16\ \text{m/s} - 9.8\ \text{m/s}^2 \times t$, so $t = 0.82\ \text{s}$.

Step 3. Coming down. $-8\ \text{m/s} = 16\ \text{m/s} - 9.8\ \text{m/s}^2 \times t$, so $t = \dfrac{24\ \text{m/s}}{9.8\ \text{m/s}^2} = 2.45\ \text{s}$.