A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01sec, the average acceleration during contace is : ` ( g=9.8 m//s ^(2)) `

Options

A. `2414 ms^(-2)`
B. `1735 ms^(-2)`
C. `3120 ms^(-2)`
D. `4105 ms^(-2)`

Answer

A

Explanation

`a_(av)=(v_(f)-v_(i))/(Delta t)=(upsilon_(f)-upsilon_(i))/(Delta t)` (as they are in opposite direction)
`=(sqrt(2gh_(f))+sqrt(2gh_(i)))/(Delta t)=(sqrt(2xx10xx5)+sqrt(2xx10xx10))/(0.01)`
`~~2414 ms^(-2)`