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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If theball is in contact with the floor for 0.01 sec, then average acceleration during contact is :-
  • a)
    2100 m/s2
  • b)
    1400 m/s2
  • c)
    700 m/s2
  • d)
    400 m/s2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A ball is dropped on the floor from a height of 10 m. It rebounds to a...
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/s^2 ~ 2100m/s^2
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Most Upvoted Answer
A ball is dropped on the floor from a height of 10 m. It rebounds to a...
2100m/s^2 , Option A.
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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If theball is in contact with the floor for 0.01 sec, then average acceleration during contact is :-a)2100 m/s2b)1400 m/s2c)700 m/s2d)400 m/s2Correct answer is option 'A'. Can you explain this answer?
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