A tennis ball is dropped on the floor from a height of 20m. It rebounds to a height of 5m. If the ball was in contact with the floor for 0.01s. What was its average acceleration during contact? `(g=10m//s^(2))`

Options

A. `3000 m//s^(2)`
B. `2000 m//s^(2)`
C. `1000 m//s^(2)`
D. `500m//s^(2)`

Answer

A

Explanation

`F_("avg")=(msqrt(2+10+10)+msqrt(2xx10xx5))/(0.01)`
`therefore a_("avg")=(10(sqrt(2)+1))/(0.01)=3000m//s^(2)`