Question

a ball falls from 20 m height on floor and rebounds to 5 m. time of contact is 0.04 sec. find acceleration during impact.

Answer 1

The top most highest potential energy = mgh.

This will be equal to the kinetic energy at the lowest point, therefore, mgh = 1/2*m*u2, or, u = √2gh = √(2*10*20) = 20 m/s.

After bouncing, the calculation will be as follows:

mgh' = 1/2*mv2 = √2gh' = √(2*10*5) = 10 m/s.

Now we know that acceleration = change in velocity / change in time

= 10-20 / 0.02

= - 500 m/s.