To test the quality of a tennis ball, you drop it onto a floor from a height of 4.00m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0ms,

(a) What is the magnitude of its average acceleration during the contact?

(b) is the acceleration up or down?

In this problem, we apply the equation regarding kinematics expressed as vf^2 = v0^2 + 2as vf eventually becomes zero because the ball stops in the end. a = - 9.8 m/s2s = 2 metres this time

This gives initial velocity, vo equal to 6.26m/s

now 6.26 - (-8.85) = 15.11m/s

change in velocity/change in time = average acceleration 15.11 / (12/1000) = 1259.167 m/s^2