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?

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Shobhit Varshney · Answer

Dear student, Velocity before collision = v 1 = &radic;(2g*10) Velocity after collision = v 2 = &radic;(2g*2.5) Acceleration = (v 1 &ndash; v 2 )/0.02 &hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.&hellip;.... Thanks

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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?

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?

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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?

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?

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