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Question

(A) Let the initial speed be $u$ and the angle of projection be $	heta$. The speed at the maximum height is $u \cos 	heta$.According to the problem,

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}$

Vedclass · Answer

(A) Let the initial speed be $u$ and the angle of projection be $	heta$. The speed at the maximum height is $u \cos 	heta$.According to the problem, the initial speed is twice the speed at the maximum height:$u = 2(u \cos 	heta)$$\Rightarrow \cos 	heta = \frac{1}{2}$$\Rightarrow 	heta = 60^{\circ}$.The range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$ and the maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.The ratio $\frac{R}{H}$ is:$\frac{R}{H} = \frac{u^2 \sin(2	heta) / g}{u^2 \sin^2 	heta / (2g)} = \frac{2 \sin(2	heta)}{\sin^2 	heta} = \frac{4 \sin 	heta \cos 	heta}{\sin^2 	heta} = 4 \cot 	heta$.Substituting $	heta = 60^{\circ}$:$\frac{R}{H} = 4 \cot(60^{\circ}) = 4 	imes \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

$A$ projectile is thrown at a speed which is twice its speed at its maximum height. If $R$ and $H$ are its range and maximum height respectively, then the ratio $\frac{R}{H}$ is

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