Tennis ball is dropped on the floor from a height of 100 m. It rebounds to a height of 100 m. Ir the ball was in contact with the floor for 3.16 3. then the average acceleration during the contact is:

Options

A. 700 `m//s^(2)`
B. 140 `m//s^(2)`
C. 280`m//s^(2)`
D. 28 `m//s^(2)`

Answer

d

Explanation

`v=sqrt(2gh) implies v=sqrt(2xx9.8xx100)=sqrt(1960)`.
After rebound, v=`-sqrt(1960)`

acceleration`=("change in velcocity")/("time")`
`(2sqrt(1960))/(3.16)=28 ms^(-2)`