## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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.