A ball is dropped on to the floor from a height of 10 m. It rebounds too a height of 2.5m.If the ball is in contact with the floor for 0.001s, what is the average accelarartion during contact?

Asked by aruni_maiyer | 8th Mar, 2011, 07:38: PM

Expert Answer:

v1 = root (2gh) = 14m/s

v2 = -7m/s.

Hence, average acceleration aaverage = v1 - (-v2)/0.02 = 21/0.02 = 1050m/s2

Answered by  | 8th Mar, 2011, 10:03: PM

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