Answer
The maximum force that the floor exerts on the ball is 936 N
Work Step by Step
We first find the ball's speed just before it hits the floor:
$v^2 = v_0^2+2gy = 0+2gy$
$v = \sqrt{2gy}$
$v = \sqrt{(2)(9.80~m/s^2)(2.0~m)}$
$v = 6.26~m/s$
We then find the ball's speed just after it rebounds from the floor;
$v_0^2 = v^2-2gy = 0-2gy$
$v_0 = \sqrt{-2gy}$
$v_0 = \sqrt{-(2)(-9.80~m/s^2)(1.5~m)}$
$v_0 = 5.42~m/s$
We can find the magnitude of the impulse that the floor exerts on the ball.
$p_0+J=p_f$
$J = p_f-p_0$
$J = m~(v_f-v_0)$
$J = (0.20~kg)[5.42~m/s-(-6.26)]$
$J = 2.34~N~s$
The impulse is equal to the area under the force versus time graph. We can use the impulse to find $F_{max}$.
$area = J$
$\frac{1}{2}F_{max}~t = J$
$F_{max} = \frac{2J}{t}$
$F_{max} = \frac{(2)(2.34~N~s)}{5\times 10^{-3}~s}$
$F_{max} = 936~N$
The maximum force that the floor exerts on the ball is 936 N.
