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The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on horizontal plane is

$\begin{array}{1 1}(A)\;\large\frac{3v^2 }{g} \\(B)\;\large\frac{3}{2} \frac{u^2}{g} \\(C)\;\large\frac{ u^2}{3g}\\(D)\;\large\frac{\sqrt 3}{2}\frac{u^2}{g} \end{array} $

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1 Answer

As $ u \cos \theta_0 =u/2 $
$\therefore \theta_o =60^{\circ}$
So $ R= \large\frac{u^2 \sin (2 \times 60^{\circ})}{g} $
$\qquad= \large\frac{\sqrt 3}{2}\frac{u^2}{g}$
Hence D is the correct answer.
answered Jul 14, 2014 by meena.p

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