The velocity at the maximum height of a proje | vedclass

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Question

(a)

$u_h=u \cos 	heta$

$\frac{\sqrt{3}}{2} u=u \cos 	heta$

$\Rightarrow \cos 	heta=\frac{\sqrt{3}}{2}$

$	heta=30^{\circ}$

$R=\frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(a)

$u_h=u \cos 	heta$

$\frac{\sqrt{3}}{2} u=u \cos 	heta$

$\Rightarrow \cos 	heta=\frac{\sqrt{3}}{2}$

$	heta=30^{\circ}$

$R=\frac{u^2 \sin 2 	heta}{g}$

$=\frac{u^2 \sin 60^{\circ}}{g} \Rightarrow \frac{\sqrt{3} u^2}{2 g}=R$

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............

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