## anonymous 5 years ago A 240 g ball is dropped from a height of 2.1 m, bounces on a hard floor, and rebounds to a height of 1.6 m. The figure shows the impulse received from the floor. What maximum force does the floor exert on the ball? impulse= .005*Fmax

$F \Delta t = m \Delta v --> F = m (\Delta v)/(\Delta t)$ We can find the speed by calculating the potential energy and converting it to kinetic energy. $E_p = E_k --> mgh = m v^2 /2$ $v = \sqrt(2gh) = \sqrt(2 * 9.81 * 2.1) \approx 6.42 m/s$ so then $F = .240 * 6.42 / 0.005=308 N$ (we know the time because of the above mentioned impulse equation.)