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

Question

anonymous · Answer

$$F \Delta t = m \Delta v  --> F = m (\Delta v)/(\Delta t)  $$ 
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.)