A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in…

A ball dropped from $h_1 = 20\text{m}$ with $g = 10\text{m/s}^2$ reaches the ground with velocity $v_{\text{before}} = -\sqrt{2gh_1} = -\sqrt{2 \times 10 \times 20} = -20\text{m/s}$.

After rebounding to $h_2 = 5\text{m}$, its upward velocity is $v_{\text{after}} = \sqrt{2gh_2} = \sqrt{2 \times 10 \times 5} = 10\text{m/s}$.

During the $\Delta t = 1\text{s}$ contact, the average acceleration is
$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_{\text{after}} - v_{\text{before}}}{\Delta t} = \frac{10 - (-20)}{1} = 30\text{m/s}^2$.

The average acceleration is $\boxed{\text{A}}$.