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.`

Explanation

Let u be velocity with which the stone is projected vertically upwards.
Given that, `v_(-h)=2v_(h)` or `(v_(-h))^(2)=4v_(h)^(2)`
`therefore u^(2)-2g(-h)=4(u^(2)-2gh)`
`therefore u^(2)=(10gh)/(3)`
Now, Maximum height `h_(max)=(u^(2))/(2g)=(5h)/(3)`