The speed of a projectile at its maximum heig | vedclass

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(C) Let the initial speed be $u$ and the angle of projection be $	heta$.The speed at maximum height is $u \cos 	heta$.Given that $u \cos 	heta = \f

Vedclass · Accepted Answer

$4\sqrt{3}$

Vedclass · Answer

(C) Let the initial speed be $u$ and the angle of projection be $	heta$.The speed at maximum height is $u \cos 	heta$.Given that $u \cos 	heta = \frac{\sqrt{3}}{2} u$,so $\cos 	heta = \frac{\sqrt{3}}{2}$,which implies $	heta = 30^{\circ}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$ and the range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.We are given $R = P 	imes H$,so $P = \frac{R}{H}$.Substituting the formulas: $P = \frac{u^2 (2 \sin 	heta \cos 	heta) / g}{u^2 \sin^2 	heta / (2g)} = \frac{4 \cos 	heta}{\sin 	heta} = 4 \cot 	heta$.For $	heta = 30^{\circ}$,$P = 4 \cot 30^{\circ} = 4 	imes \sqrt{3} = 4\sqrt{3}$.

The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it,$P$ is equal to

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