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?

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

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.

To determine the average acceleration of the tennis ball during contact with the floor, you can use the equation:

average acceleration = (final velocity - initial velocity) / time

In this case, the initial velocity is the velocity of the tennis ball just before it hits the floor. Since it is dropped from a height of 4.31 m, we can use the formula to calculate the initial velocity. The formula to determine the velocity of an object in free fall is:

v = √(2gh)

Where:
v = velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
h = height

Plugging in the values, we get:

v = √(2 * 9.8 * 4.31)
v ≈ 8.84 m/s

Next, the final velocity is the velocity of the ball just after it rebounds from the floor. Since it reaches a height of 2.51 m, we can again use the formula to calculate the final velocity. Rearranging the formula, we have:

v = √(2gh)

Plugging in the values, we get:

2.51 = 0 + √(2 * 9.8 * h)

Solving for h, we get:

h ≈ 0.332 m

Now, we can calculate the final velocity using this value:

v = √(2 * 9.8 * 0.332)
v ≈ 2.83 m/s

Substituting the values into the average acceleration equation:

average acceleration = (2.83 - 8.84) / 0.0126
average acceleration ≈ -500.4 m/s²

Therefore, the average acceleration of the tennis ball during contact with the floor is approximately -500.4 m/s². The negative sign indicates that the acceleration is directed opposite to the initial velocity.