Answer
The magnitude of the average acceleration is $~~1.26\times 10^3~m/s^2$
Work Step by Step
We can find the velocity when the ball reaches the floor:
$v^2 = v_0^2+2ay$
$v^2 = (0)^2+(2)(-9.8~m/s^2)(-4.00~m)$
$v = \sqrt{78.4}$
$v = -8.85~m/s$
We can find the velocity $v_0$ when the ball leaves the floor:
$v^2 = v_0^2+2ay$
$v_0^2 = v^2-2ay$
$v_0^2 = (0)^2-(2)(-9.8~m/s^2)(2.0~m)$
$v_0 = \sqrt{39.2}$
$v_0 = 6.26~m/s$
For the next part of the question, let $v_0 = -8.85~m/s$ and let $v = 6.26~m/s$
We can find the acceleration:
$v = v_0+at$
$a = \frac{v - v_0}{t}$
$a = \frac{(6.26~m/s) - (-8.85~m/s)}{12.0\times 10^{-3}~s}$
$a = 1.26\times 10^3~m/s^2$
The magnitude of the average acceleration is $~~1.26\times 10^3~m/s^2$.
