A stone is dropped from the top of a vertical pillar. When the stone has fallen through a height x, another stone is dropped from height y below the top of the pillar. Both the stones touch the ground at the same time . Prove that the height of the pillar should `(x+y)^(2)/(4x)`.

Explanation

Let the hight of the pillar be h and the velocity of the stone at x below the lop of the pillar be v. `:. V^(2) = 2gx " ""or", v = sqrt(2gx) " "cdots(1)`
Let the stone take t s to cover the distance (h-x) .
`:." " h-x = vt + (1)/(2)"gt"^(2) " "cdots(2)`
According to the problem the stone is dropped from a height of (h-y) and this stone takes time t to cross the distance.
`:. " " h-y = (1)/(2) "gt"^(2)`
From equations (2) and (3) we get,
h-x = vt +h-y`" ""or", y -x = vt " ""or" , y - x = t sqrt(2gx)`
`:. " " t = (y-x)/(sqrt(2gx))`
From equation (3) we get ,
`h-y = (1)/(2) g ""(y-x)^(2)/(2gx) = (y-x)^(2)/(4x)`
or, `" " (h-y). 4x = (y-x)^(2) " ""or",4xh - 4xy = y^(2) -2yx +x^(2)`
or, `" " h = ((x+y)^(2))/(4x)` (Proved).