To test the quality of a tennis ball, you drop it onto the floor from a height of 4.31 m. It rebounds to a height of 2.51 m. If the ball is in contact with the floor for 12.6 ms, what is its average acceleration during that contact?

Answer 1

V1^2 = Vo^2 + 2g*d
V1^2 = 0 + 19.6*4.31 = 84.48
V1 = 9.2 m/s.

V^2 = Vo^2 + 2g*d = 0
Vo^2 + (-19.6)*2.31 = 0
Vo^2 = 45.3
Vo = 6.73 m/s.

a =(Vo-V1)/t
a=(6.73-9.2)/12,6=-0.196m/s^2

Answer 2

CORRECTION: Change 2.31 io 2.51 m.
Vo^2 = 49.2
Vo = 7.01 m/s, Up ward.

a = (-7.01-9.2)/12.6 = -1.29 m/s^2.