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(D) $1$. For option $A$: Maximum height $H = u^2 / (2g)$ when $	heta = 90^\circ$. Maximum range $R_{max} = u^2 / g$ when $	heta = 45^\circ$. Thus,$R

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All of the above.

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(D) $1$. For option $A$: Maximum height $H = u^2 / (2g)$ when $	heta = 90^\circ$. Maximum range $R_{max} = u^2 / g$ when $	heta = 45^\circ$. Thus,$R_{max} = 2H$. Option $A$ is correct.$2$. For option $B$: $R = (u^2 \sin 2	heta) / g$ and $H = (u^2 \sin^2 	heta) / (2g)$. Given $R = nH$,we have $(2u^2 \sin 	heta \cos 	heta) / g = n \cdot (u^2 \sin^2 	heta) / (2g)$. This simplifies to $4 \cos 	heta = n \sin 	heta$,so $	an 	heta = 4/n$ or $	heta = 	an^{-1}(4/n)$. Option $B$ is correct.$3$. For option $C$: $T = (2u \sin 	heta) / g$,so $T^2 = (4u^2 \sin^2 	heta) / g^2$. Also $R = (u^2 \sin 2	heta) / g = (2u^2 \sin 	heta \cos 	heta) / g$. Then $2R 	an 	heta = 2 \cdot [(2u^2 \sin 	heta \cos 	heta) / g] \cdot [\sin 	heta / \cos 	heta] = (4u^2 \sin^2 	heta) / g = gT^2$. Option $C$ is correct.Since all statements are correct,the answer is $D$.

Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

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