A ball is dropped on the floor a height of `80 m` rebounds to a height of `20 m`. If the ball is in contact with floor for 0.1 s, the average acceleration during contact is

Options

A. `400 m//s^(2)`
B. `500 m//s^(2)`
C. `600 m//s^(2) `
D. `800 m//s^(2)`

Answer

C

Explanation

`A` to `B`: `u^(2)=0+2gxx80`
`impliesu=40 m//s`
`B` to `C: 0=v^(2)-2gxx20`
`impliesv=20 m//s`
`bar(a)=(Deltav)/(Deltat)=(v+u)/(Deltat)=(40+20)/0.1=600 m//s^(2)`