A stone is projected horizontally with speed v from a height h above ground. A horizontal wind is blowing in direction opposite to velocity of projection and gives the stone a constant horizontal acceleration f( in a direction opposite to initial velocity ) .As a result the stone falls on ground at a point vertically below the point of projection . Then the value of height h in terms of f, g,v is (g is acceleration due to gravity)

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$\frac{g v^{2}}{2 f^{2}}$

$\frac{\sqrt{2} g v^{2}}{f^{2}}$

$\frac{g v^{2}}{f^{2}}$

$\frac{2 g v^{2}}{f^{2}}$

answer is A.

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Detailed Solution

Then taken to reach the ground is given by

$h = \frac{1}{2} g t^{2}$ …..(1)

Since horizontal displacement in time t is zero

$∴ t = \frac{2 v}{f}$ …..(2)

From eq .(1) and (2)

$h = \frac{2 g v^{2}}{f^{2}}$

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