To test the quality of a tennis ball, you drop it onto the floor from a height of 4.23 m. it rebounds to a height of 2.54 m. if the ball is in contact with the floor for 12.2 ms, what is its average acceleration during that contact?

Step-by-step explanation:

The average acceleration is given by:

where is the velocity of the ball an instant before hitting the ground and an instant after.

we can obtain by applying the following formula:

where since the ball falls from rest, g is the acceleration of gravity and

then:

(note that here we select the negative result because of the downwards direction of the velocity)

Similarly, we calculate with the formula:

where is the velocity at the maximum height, and

then:

Finally the average acceleration is:

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)