A ball is dropped from a height of 10 m and rebounds to a height of 2.5 m. If it is in contact with the floor for 0.01 s, what is the average acceleration during contact?

What you are really finding during the bounce

During the very short time the ball touches the floor, its velocity reverses direction, from downward to upward. Average acceleration during contact is based on how quickly that velocity change happens.

Speed just before hitting the floor (from 10 m)

Dropping from rest from height $h_1=10\,\text{m}$ gives $$v_1 = \sqrt{2gh_1} = \sqrt{2(9.8)(10)} = 14\,\text{m/s}.$$ Direction is downward, so with upward positive: $v_{\text{before}}=-14\,\text{m/s}$.

Speed just after leaving the floor (to reach 2.5 m)

To rebound up to $h_2=2.5\,\text{m}$, the launch speed upward must be $$v_2 = \sqrt{2gh_2} = \sqrt{2(9.8)(2.5)} = 7\,\text{m/s}.$$ So $v_{\text{after}}=+7\,\text{m/s}$.

Average acceleration over the 0.01 s contact time

Compute the change in velocity: $$\Delta v = v_{\text{after}} - v_{\text{before}} = 7 - (-14) = 21\,\text{m/s}.$$ Then $$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100\,\text{m/s}^2.$$ The positive sign means the average acceleration is upward.