In the projectile motion of an object,the obj | vedclass

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(C) At the maximum height,the vertical component of velocity is zero,so the speed of the projectile is $v = u \cos 	heta$. Given that the speed at ma

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

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(C) At the maximum height,the vertical component of velocity is zero,so the speed of the projectile is $v = u \cos 	heta$. Given that the speed at maximum height is half of the initial speed $u$,we have $u \cos 	heta = \frac{1}{2} u$,which implies $\cos 	heta = \frac{1}{2}$,so $	heta = 60^{\circ}$. The range $R$ of a projectile 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 of range to maximum height is $\frac{R}{H} = \frac{u^2 \sin 2 	heta / g}{u^2 \sin^2 	heta / 2g} = \frac{2 \sin 2 	heta}{\sin^2 	heta}$. Using the identity $\sin 2 	heta = 2 \sin 	heta \cos 	heta$,we get $\frac{R}{H} = \frac{2(2 \sin 	heta \cos 	heta)}{\sin^2 	heta} = 4 \cot 	heta$. Substituting $	heta = 60^{\circ}$,we get $\frac{R}{H} = 4 \cot 60^{\circ} = 4 	imes \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

Column $I$	Column $II$
$(A)$ $u_x$ is doubled, $u_y$ is halved	$(p)$ $H$ will remain unchanged
$(B)$ $u_y$ is doubled, $u_x$ is halved	$(q)$ $R$ will remain unchanged
$(C)$ $u_x$ and $u_y$ both are doubled	$(r)$ $R$ will become four times
$(D)$ Only $u_y$ is doubled	$(s)$ $H$ will become four times

In the projectile motion of an object,the object reaches its maximum height where its speed is half of its initial speed. Then the ratio between the range and the maximum height of the projectile is

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