A ball is dropped on to the floor from a height of 10m. It rebounds to a height of 2.5m.If the ball is in contact with the floor for 0.01s, what is the average acceleration during contact?

Asked by  | 25th Sep, 2012, 08:31: PM

Expert Answer:

velocity of the ball just before contact:
u = -(20g)1/2    using energy conservarion, i.e equating PE at top with KE at bottom
velocity of ball just before leaving ground:
v = (5g)1/2
thus force imparted to ground :
F = M ( (5g)1/2  + (20g)1/2   )/0.01  the direction of force is downwards.

Answered by  | 27th Sep, 2012, 09:23: AM

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