A rubber ball of mass 100 g falls from a height of 1m and rebounds to a height of 40 cm.Find the impulse and the average force between the ball and the ground, if time during which they are in contact was 0.1 sec.

While falling down, initial velocity = 0, final velocity with which the body strikes the ground = v m/s, acceleration due to gravity = 9.8 m/s ^{ 2 }, height = 1m

The velocity of the ball with which it hits the ground can be found using,

v ^{ 2 }= u ^{ 2 }+ 2as

=> v ^{ 2 }= 0 + 2×9.8×1

=> v = 4.43 m/s (downward)

While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s ^{ 2 }, height = 0.4 m

The velocity of the ball with which it leaves the ground can be found using,

v ^{ 2 }= u ^{ 2 }+ 2as

=> 0 = u ^{ 2 }- 2×9.8×0.4 [this time the velocity and acceleration are oppositely directed so we have the negative sign]

=> u =2.8 m/s (upward)

So,Magnitude of the change in velocity = 4.43-2.8=1.63 m/s

=> Δv = 1.63 m/s

∴ Impulse=change in linear momentum,dp = mΔv = 100 x 1.63=163 kg m/s

So, Force because of acceleration achieved during the strike is, F = dp/dt = 163/0.1 = 1630 N

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