Question

a tennis ball is dropped on to the floor from a height of 4 m. It rebpunds to a height of 3 m. If the ball was in contact with the floor for 0.01 s, what was its average acceleration during contact?

Answer 1

u = sqrt(2gH)
= -sqrt(2 × 9.8 × 4)

v = -sqrt(2gh)
= sqrt(2 × 9.8 × 3)

Average acceleration = (v - u) / t
= (v - (-u)) / t
= (v + u) / t
= [sqrt(2 × 9.8 × 3) + sqrt(2 × 9.8 × 4)] / 0.01
= 1652.25 m/s^2

Average acceleration during contact is 1652.25 m/s^2

Answer 2

Answer:

Amswer is given above.