A table tennis ball is dropped unto the floor from a height of 4m and it rebounds to a height of 3m. If the time of contact with the floor is 0.01secs, what is the magnitude and direction of the acceleration during the contact

Answer 1

h = 0.5g*t^2 = 4 m.
4.9*t^2 = 4
t^2 = 0.816
Tf = 0.904 s. = Fall time.

V1^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4
V1 = 8.85 m/s., Downward.


V^2 = V2^2 + 2g*h = 0
V2^2 = -2g*h = -2*(-9.8)*3 = 58.8
V2 = 7.67 m/s, Upward.

a = (V2-V1)/t = (7.67-8.85)/0.01 = -118
m/s^2, Upward.

Answer 2

Note: The Fall time is not required.