The velocity at the maximum height of a proje | vedclass

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Question

$u\cos 	heta = \frac{u}{2}$

$	herefore \,\,	heta = 60^\circ $

Now $R = \frac{{{u^2}\sin (2 	imes 60^\circ )}}{g}$

$R = \frac{{\sqrt 3 }}{

Vedclass · Accepted Answer

$\frac{\sqrt{3} u ^{2}}{2 g }$

Vedclass · Answer

$u\cos 	heta = \frac{u}{2}$

$	herefore \,\,	heta = 60^\circ $

Now $R = \frac{{{u^2}\sin (2 	imes 60^\circ )}}{g}$

$R = \frac{{\sqrt 3 }}{2}\frac{{{u^2}}}{g}$

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

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