1. Find the velocity just before hitting the floor

The ball is dropped from rest from height $h_1 = 9.8\,\text{m}$.

Using $v^2 = u^2 + 2gh$ with $u=0$, $v_1 = \sqrt{2gh_1} = \sqrt{2 \cdot 10 \cdot 9.8} = \sqrt{196} = 14\,\text{m/s}$

This velocity is downward. Taking upward as positive: $v_i = -14\,\text{m/s}$

2. Find the velocity just after rebound

The ball rises to height $h_2 = 5.0\,\text{m}$.

Again, $v^2 = u^2 + 2as$ At the top, final velocity is $0$, acceleration is $-g$, and displacement is $+5$ m. So rebound speed is $v_2 = \sqrt{2gh_2} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10\,\text{m/s}$

This is upward, so $v_f = +10\,\text{m/s}$

3. Compute change in velocity during contact

$\Delta v = v_f - v_i = 10 - (-14) = 24\,\text{m/s}$

4. Find average acceleration

Time of contact: $\Delta t = 0.2\,\text{s}$

Therefore, $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120\,\text{m/s}^2$

5. Final answer

The average acceleration during contact is $\boxed{120\,\text{m/s}^2}$

This matches the stored correct answer.