A ball is dropped on to the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, what is the average acceleration during contact ?

Options

A. `700 m s^(-2)`
B. `1400 m s^(-2)`
C. `2100 m s^(-2)`
D. `2800 m s^(-2)`

Answer

C

Explanation

`v_(1)=sqrt(2gh)=sqrt(2xx10xx10)=sqrt(200)`
`v_(2)=-sqrt(2gh)=-sqrt(2xx10xx2.5)=-sqrt(50)`
So `Deltav=v_(1) -v_(2) =sqrt(200) + sqrt(50) =3sqrt(50) =21 ms^(-1)`
`:.` Average acceleration `=(Deltav)/(Deltat) = 21/(0.01) = 2100 ms^(-2)`