Question

A ball is dropped on the floor from a height of 10 m rebounds to a height
of 2.5 m. If the ball is in contact with the floor for 0.02 sec, its average
acceleration during contact is​

Answer 1

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When it is dropped from 10m,

Initial height = 10m

initial velocity = 0

velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)

after rebound,

maximum height reached = 2.5m

final velocity at top = 0

initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)

assuming downward as positive direction

So velocity just before hitting ground = +14.07 m/s 

velocity just after hitting ground = -7 m/s 

change in velocity = +14.07 - (-7) = 21.07 m/s

time = 0.01s

acceleration = change in velocity/time = 21.07/0.01 = 2107 m/s² ~ 2100m/s²

Answer 2

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