Physics ball problem

A tennis ball is dropped from 1.71mabove the
ground. It rebounds to a height of 0.874 m.
acceleration of gravity=9.8 m/s^2 (down is negative)

If the tennis ball were in contact with the
ground for 0.0127 s, find the acceleration
given to the tennis ball by the ground.
Answer in units of m/s^2



is this some trick question, i thought the acceleration would always be 9.8?...

Yes, the acceleration due to gravity at the surface of the earth is approx 9.8 m/s^2 -- always. But they're asking for the acceleration that the collision with the ground gives the ball. There is a large change in speed during the 12.7 millisecs of the bounce. Gravity would not cause near that change in speed in that short a time.

You need the speed just before the contact with the ground. Use the kinematic formula
Vf^2 = Vi^2 + 2*a*y
where Vi = 0, a = -9.8 m/s^2, and y is the distance it falls or -1.71 m.

And you need the speed just after contact with the ground. Use the same formula where
Vf = 0 (it comes to a stop at the top), a = -9.8 m/s^2, and y = 0.874 m.

Now use the kinematic formula
Vf = Vi + a*t
where Vf is the speed just after the bounce, Vi is the speed just before the bounce, and t = 0.0127 s. Solve for a.