Q. A body is thrown horizontally with a velocity u from the top of a tower. The displacement of the stone when the horizontal and vertical velocities are equal is

(a) $ \displaystyle \frac{u^2}{g}$

(b) $ \displaystyle \frac{u^2}{2 g}$

(c) $\displaystyle \frac{\sqrt5 u^2}{2g}$

(d) $ \displaystyle \frac{2 u^2}{g}$

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Ans: (c)

Sol: $ v_x = v_y$

$ u = g t $

$t = \frac{u}{g} $ … (i) ; (since horizontal velocity remains unchanged )

After time t , horizontal distance , $x = u t = \frac{u^2}{g}$

After time t , vertical distance , $ y = \frac{1}{2}g t^2 = \frac{u^2}{2 g}$

Displacement $= \sqrt{x^2 + y^2 }$

$= \sqrt{(\frac{u^2}{g})^2 + (\frac{u^2}{2g})^2}$

$= \frac{\sqrt{5}}{2} \frac{u^2}{g}$