A ball is dropped on the floor from the height of 10m. It rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01 seconds, what is the average acceleration during contact ?

Here,

Height from which the ball is dropped is, h = 10 m

Velocity with which the ball hits the ground can be found as,

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

=> v^{2} = 2gh

=> v = (2gh)^{1/2} [downward]

The ball then rebounds to a height of, h^{/} = 2.5 m. Let the velocity with which the ball rebounds be v^{/}.

So, using,

0^{2} = (v^{/})^{2} – 2gh^{/}

=> v^{/} = (2gh^{/})^{1/2} [upward]

The time for which the ball was in touch with the ground is t = 0.01 s

So, acceleration of the ball is,

a = (v^{/} - v)/t

[considering upward velocity to be positive, v^{/} is positive and v is negative]

=> a = [(2gh^{/})^{1/2} – {-(2gh)^{1/2}}]/t

=> a = [(2×9.8×2.5)^{1/2} + (2×9.8×10)^{1/2}]/0.01

=> a = 2100 m/s^{2}

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