A ball is dropped on the floor from a he...

A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5 m if the ball is in contact with floor for 0.01 s then the average acceleration during contact is nearly

`2100"m/s"^(2)` downward

`2100"m/s"^(2)` upward

`1400"m/s"^(2)`

`700"m/s^(2)`

Text Solution

Verified by Experts

The correct Answer is:

Velocity when ball strikes the ground `V=sqrt(2gh_(1))`
`V=sqrt(2xx10xx10)=sqrt(200)`
Velocity of ball after rebound `V=sqrt(2gh_(2))`
`V=sqrt(2xx10xx2.5)=sqrt(50)`
Change in velocity /time
`=(sqrt(50)-(-sqrt(200)))/(0.01)=(7.07+14.114)/(0.01)=2121.2"m/s"^(2)`

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