A ball of mass 1kg dropped from 9.8m height strikes the ground and rebounds to a height of 4.9m. If the time of contact between ball and ground is 0.1s, then find impulse and average force acting on ball.

v2 = u2 + 2gh

v2 = 0 + 2 x 9.8 x 9.8

v = √(2 x 9.8 x 9.8)

We get,

v = 9.8 √2

Again, v2 = u2 – 2gh

0 = u2 – 2 x 9.8 x 4.9

u2 = 96.04

u = √96.04 = 9.8

So,

u = 9.8

Impulse = m[v-(-u)]

= 1 [9.8 √2 – (-9.8)]

= 1 [9.8 √2 + 9.8]

= 9.8 (√2 + 1)

= 9.8 x 2.4

We get,

= 23.52 N – s

We know that,

Force = Impulse / Time

Force = 23.52 / 0.1

We get,

Force = 235.2 N

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