A tennis ball is dropped on the floor from a height of 20 rebounds to a height of 5 metre the ball was in contact with the floor for 0.101 seconds
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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
= -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
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