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,

v² = u² + 2gh

=> v² = 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,

0² = (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/s²