When a stone falling from the top of vertical tower has fallen a distance x m, another is let fall from a point y m below the top. If the fall from rest and reach the ground together, then the height of the tower is

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$\frac{(x + y)^{2}}{4 x} m$

$4 x (x + y)^{2} m$

$\frac{4 (x + y)^{2}}{x} m$

$\frac{4 x}{(x + y)^{2}} m$

answer is A.

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Detailed Solution

Let the height of tower be h,

Time taken by first stone to reach ground from h-x is equal to the time taken by second stone to reach ground from height h-y,

$\begin{array}{l} \sqrt{\frac{2 h}{g}} - \sqrt{\frac{2 x}{g}} = \sqrt{\frac{2 (h - y)}{g}} \\ \Rightarrow \sqrt{h} - \sqrt{x} = \sqrt{h - y} \\ h + x - 2 \sqrt{hx} = h - y; x + y = 2 \sqrt{hx} \\ h = \frac{(x + y)^{2}}{4 x} \end{array}$

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