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

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