Question

A ball of mass 1 kg is dropped from 9.8 m height strikes the ground and rebiunds 4.9 m. if the time of contact between ball and ground id 0.1 s, then find impluse andnaverage forve acting on the ball...justify your answer

Answer 1

Velocity when it hits the ground
u = sqrt(2gH)
= sqrt(2 * 9.8 * 9.8)
= 9.8 sqrt(2) m/s

Velocity just after hitting the ground
v = sqrt(2gH)
= sqrt(2 * 9.8 * 4.9)
= 9.8 m/s

Impulse = change in momentum
= mass × change in velocity
= m * (v - u)
= 1kg * (9.8 - (-9.8*sqrt(2)) m/s
= 9.8 * (1 + sqrt(2)) kgm/s
= 9.8 * 1.414 kgm/s
= 13.8572 kgm/s

Average Force = Impulse / Time taken
= 13.8572 / 0.1
= 1.38 N

Answer 2

Answer:

The impulse will be 23.61 N-s

And Average force will be 236 N.

For detailed explanation plz go through the image.....

Thank you.

Hope it helps....

Explanation: