# 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 a

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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.
A. (3)/(5) h
B. (5)/(3) h
C. (7)/(5) h
D. (5)/(7) h

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At height h above P:
v^(2) = u^(2) - 2gh ....(i)
At depth h below P:
(2v)^(2) = u^(2) + 2gh....(ii)
5v^(2) = 2u^(2)
u^(2) = (5)/(2) v^(2)
Eqn. (ii) minus Eqn. (i) gives
3v^(2) = 4 g h rArr v^(2) = (4)/(3) gh
The maximum height attained by the stone is
H = (u^(2))/(2g) = (5v^(2)//2)/(2g) = (5v^(2))/(4g) = (5)/(4g) ((4)/(3) gh) = (5)/(3) h