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 $v_1$ be the speed of the ball just before it hits the ground, and $v_2$ be the speed just after.

Using energy:

$$mg(5) = \frac{1}{2} m v_1^2$$
$$10g = v_1^2$$
$$v_1 = \sqrt{10g}$$

Also:
$$mg(3.2) = \frac{1}{2} m v_2^2$$
$$6.4g = v_2^2$$
$$v_2 = \sqrt{6.4g}$$

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

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

 

[DR33]

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

l8er