From an elevated point 'p' a stone is projected vertically upwards. When it reaches a distance 'd' below 'p' its velocity is doubled.The greatest height reaches by it above 'p' is

Asked by sowmya subathra | 17th Apr, 2013, 08:30: PM

Expert Answer:

Ans1.



Let initial velocity = u
Therefore at point 'P', AP = h
VP2 - u2 = -2gh --------(1)
At point Q, VQ2 - u2 = 2 (-g)(-h) = +2gh
As VQ = 2Vp, so
4VP2 - u2 = 2gh -------(2)
4[u2 - 2gh] - u2 = 2gh
4u2 - 8gh - u2 = 2gh
3u2 = 10 gh
u2 = (10/3)gh -------(3)
But 02 - u2 = -2gH ------(4)
\ 2gH = (10/3)gh
H = (5/3)h

Answered by  | 18th Apr, 2013, 03:06: PM

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