The velocity at the maximum height of a proje | vedclass

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Question

(A) At the maximum height,the vertical component of velocity is zero,so the velocity of the projectile is equal to its horizontal component,$v = u \co

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) At the maximum height,the vertical component of velocity is zero,so the velocity of the projectile is equal to its horizontal component,$v = u \cos 	heta$.Given that the velocity at the maximum height is $\frac{\sqrt{3}}{2}$ times the initial velocity $u$,we have:$u \cos 	heta = \frac{\sqrt{3}}{2} u$$\cos 	heta = \frac{\sqrt{3}}{2}$$	heta = 30^{\circ}$The horizontal range $R$ of a projectile is given by the formula:$R = \frac{u^2 \sin 2	heta}{g}$Substituting $	heta = 30^{\circ}$ into the formula:$R = \frac{u^2 \sin(2 	imes 30^{\circ})}{g} = \frac{u^2 \sin 60^{\circ}}{g}$Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get:$R = \frac{u^2 	imes \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3} u^2}{2 g}$Thus,the correct option is $A$.

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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