Answer to Question #113839 in Classical Mechanics for Caylin

Question

A 200g ball is dropped from a height of 2.0m, bounces on a hard floor and rebounds to a height of 1.5m and the interval for total travel is t= 5.0 ms.

What maximum force(Fmax) does the floor exert on the ball?

Please assist

Answer 1

The impulse of a ball when it rebounds from a hard floor

$$\Delta p=|m{\bf v}_f-m{\bf v}_i|=m(v_f+v_i)\\ =m\left(\sqrt{2gh_f}+\sqrt{2gh_f}\right)\\ =0.2\left(\sqrt{2\times 9.8\times 2.0}+\sqrt{2\times 9.8\times 1.5}\right)=2.3\:\rm N\cdot s$$

The force that floor exert on the ball

$$F=\frac{\Delta p}{\Delta t}=\frac{2.3}{0.005}=460\:\rm N$$