Question

a ball of mass 1kg dropped from height 9.8 m, strikes with ground and rebounds at height of 4.9 m, if the tym of contact between ball and ground is 0.1 s then find impulse and average force acting on ball​

Answer 1

Answer:

Okay :)

m = 1 kg.

Impulse = force × time.

h = 9.8 m.

h1 = 4.9 m.

v = √(2×9.8×9.8)

= 9.8√2

v2 = √(9.8 × 9.8)

= 9.8

Change in velocity = 9.8(√2 - 1)

Force = change in momentum/time

= 1 × 9.8(√2 - 1)/0.1

= 98(√2 - 1) N

Impulse = force × time

= 9.8(√2 - 1) Ns