To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact? Take `g= 98 m//s^2.`

Answer

A, B, C, D

Explanation

`a_(av)=(Deltav)/(Deltat)=(v_f-v_i)/(Deltat)`
`=(sqrt(2gh_f)+sqrt(2gh_i))/(Deltat)`
`=(sqrt(2xx9.8xx2)+sqrt(2xx9.8xx4))/(12xx10^-3)`
`=1.26xx10^3 m//s^2`