Question

A 160 g ball is dropped from a height of 2.2 m , bounces on a hard floor, and rebounds to a height of 1.1 m . The figure(Figure 1) shows the impulse received from the floor.

**Part A**

What maximum force does the floor exert on the ball?

Answer #1

Get v when it hits the ground = sqrt(2gh) = 6.57 m/s

When it rebounds speed = Sqrt(2*9.8*1.1) = 4.64 m/s

Change in momentum = 0.16 kg ( 4.64-(-6.57) = 1.794 kgm/s

Ohhh figure one is needed after this

Any way from the figure get the time interval actully if Force is constant , Or if not constant get the area under the graph. And make that area equal to 1.794 , To get Force !

Remember Force*time (Area under force-time graph)= impulse=1.794

post with figure ,if you don't get the answer

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