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From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is doubleof what it was at a height h above A. Show that the greatest height attained by the stone is `5/3 h.`

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Correct Answer - A
Let u be the velocity with which the stone is projected vertically upwards.
Given that, `v_(-h)=2v_h`
or `v_(h)^2=4v_h^2`
`:. u^2-2g(-h)=4(u^2-2gh)`
`:. u^2=(10gh)/3`
Now, `h_(max)=u^2/(2g)=(5h)/3` Hence proved.

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