Answer
857.5 m/s^2
Work Step by Step
We first find how fast the ball is falling before it hits the ground:
$v_f^2 = v_i^2-2ad$
$v_f = \sqrt{0-2*9.8m/s^2*(-15m)} = -17.15m/s$
The average acceleration is equal to the change in velocity over the change in time. Thus, we obtain:
$a = (17.15m/s)/(0.02s) = 857.5 m/s^2$
