A ball is falling freely from a certain height. When it reached `10 m` height from the ground its velocity is `v_(0)`. It collides with the horizontal ground and loses `50 %` of its energy and rises back to height of `10 m`. The value of velocity `v_(0)` isA. 7 m/sB. 10 m/sC. 14 m/sD. 16 m/s

A ball is falling freely from a certain height. When it reached `10 m`

1.	A ball is falling freely from a certain height. When it reached `10 m` height from the ground its velocity is `v_(0)`. It collides with the horizontal ground and loses `50 %` of its energy and rises back to height of `10 m`. The value of velocity `v_(0)` isA. 7 m/sB. 10 m/sC. 14 m/sD. 16 m/s
Answer» Correct Answer - C The velocity of the body on reaching the ground is `V = sqrt(u^(2) + 2gh) = sqrt(V_(0)^(2) + 196)` Given that, `(1)/(2) (KE) = PE` `(1)/(2) ((1)/(2) mV^(2)) = mgh` `V^(2) = 4 gh` `V_(0)^(2) + 196 = 4 xx 9.8 xx 10` `V_(0)^(2) = 392 - 196 = 196` `V_(0) = 14 m//s`

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A ball is falling freely from a certain height. When it reached `10 m` height from the ground its velocity is `v_(0)`. It collides with the horizontal ground and loses `50 %` of its energy and rises back to height of `10 m`. The value of velocity `v_(0)` isA. 7 m/sB. 10 m/sC. 14 m/sD. 16 m/s

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