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a ball is dropped on to the floor from a height of 10m rebounces to a height 2.5m.if the ball is in contact with the floor for 0.02s,its average acceleration during contact is?

nithin , 10 Years ago
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anser 1 Answers
Shobhit Varshney

Last Activity: 10 Years ago

Dear student,

Velocity before collision = v1 = √(2g*10)
Velocity after collision = v2 = √(2g*2.5)

Acceleration = (v1 – v2)/0.02
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Thanks
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