To test the quality of a tennis ball you drop it onto the floor from a height of 4.00m. It rebounds to...?

Question

a height of 3.00m. If the ball was in contact with the floor for 10.0ms, what was its average acceleration during contact?

Answer 1

First calculate the velocity of the ball right before it hits the floor:

Vf^2 = 2ad = 2*(-9.8m/s^2)*(-4m) =

Vf = 8.85 m/s downward (or -8.85 m/s)

Now calculate the velocity needed to get to 3.0m in height.

Vf^2 = Vo^2 + 2 a d

0m/s = Vo^2 + 2 (-9.8m/s^2)(3m)

Vo = 7.67 m/s upwards ( or +7.67 m/s)

After it hits it goes from -8.85m/s to +7.67m/s in 0.01s

a_avg = (Vstop - Vinitial) / t = (7.67m/s - (-8.85 m/s))/ 0.01 s

ANS: a_avg =  1652 m/s^2

Hope that helps.