a ball is dropped on floor from height of 10m .If the ball is in contact with floor for 0.01 s , then the average acceleration during contact is nearly_______________
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(v-u) /t
(10-0)/0.01=10/0.01
=1000m/s square
(10-0)/0.01=10/0.01
=1000m/s square
rajasekharironman:
sorry its wrong
(vf-vi) /(tf-ti) =average acceleration
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