What's the acceleration when it contacts the floor?

[DR33]

A tennis ball is dropped from 5m High and bounces back 3.2 m high. It stays in contact with the floor for 0.036 seconds. What's the acceleration when it contacts the floor?

The answer is 495 m/s_². I just don't know how to get to it. thanks so much.

 

[Tron3k]

Let [itex]v_1[/itex] be the speed of the ball just before it hits the ground, and [itex]v_2[/itex] be the speed just after.

Using energy:

[tex]mg(5) = \frac{1}{2} m v_1^2[/tex]
[tex] 10g = v_1^2 [/tex]
[tex] v_1 = \sqrt{10g} [/tex]

Also:
[tex]mg(3.2) = \frac{1}{2} m v_2^2[/tex]
[tex]6.4g = v_2^2 [/tex]
[tex]v_2 = \sqrt{6.4g} [/tex]

Acceleration is the change in velocity divided by time. (Now consider the up direction to be positive)

[tex] a = \frac{\Delta v}{\Delta t}[/tex]
[tex] a = \frac{\sqrt{6.4g} - (-\sqrt{10g})}{0.036}[/tex]
[tex] a = 495 m/s^2[/tex]

 

[DR33]

Thx so much.. I really appreciate people helping on other'S request.
thx again

l8er