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A ball dropped on to 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 0.02s, its average acceleration during contact is

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By Expert Faculty of Sri Chaitanya
a
2100 ms-2
b
1050 ms-2
c
4200 ms-2
d
9.8 ms-2

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detailed solution

Correct option is B

The velocities of the ball before and after contact with ground are 

v = 2gh1 and u =- 2gh2  (assuming downward quantities are positive) 

a=vut=2gh2+2gh1t=2gh2+h1t=2×9.8(2.5+10)0.02=1050m/s2


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