a golf ball released from a height of 1.5 m above aa concreate floor bounces back to a height of 1.1m if the ball is in contact with the floor for .62 ms what is the average acceleartion of the ball while in contact with the floor?

Answer 1

Vf^2=2ad

so in each of these, 1.5 meter, 1.1m find the floor velocity involved. Notice that they are opposite directions. a is g.

Now, having those two velocities, V1, V2

a=(V2-V1)/t but working in just speeds, and knowing they are in opposite directions

a=(V2+V1)/t