A ball of mass m1= 0.08 kg starts from rest and falls vertically downward from a height 3 m. After colliding with the ground, it bounces up to a...

We can use equations of motion to solve this problem. Using 

for the part where the ball drops from the height of 3 m, we can calculate the ball's velocity as it hits the ground. Here, u = 0, a = g = 9.81 m/s^2 and s = 3 m.

Thus, v^2 = 0^2 + 2 x 9.81 x 3

solving the equation, we get, v = 7.67 m/s.

Thus, momentum of the ball immediately before collision = mv

= 0.08 kg x 7.67 m/s = 0.61 kg m/s.

After the impact, the ball rises up to a height of 2 m. Using the same equation of motion, we know that v = 0 m/s, s = 2 m and a = -g = -9.81 m/s^2

we can find u by: u^2 = v^2 - 2as = 0 - 2 x (-9.81) x 2 

solving this, we get, u = 6.26 m/s.

Thus the momentum of the ball immediately after collision with the ground is 

= m x u = 0.08 kg x 6.26 m/s = 0.50 kg m/s.

Force can be calculated as the rate of change of momentum. 

Thus, the average force exerted = (0.61 - 0.5)/0.005 = 22 N.

The impulse is nothing but the change in momentum and is equal to 0.11 kg m/s.

Hope this helps.