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(C) For a projectile projected at an angle $	heta$ with the horizontal:Maximum height,$H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$Range,$R = \frac{u^2 \

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$2$

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(C) For a projectile projected at an angle $	heta$ with the horizontal:Maximum height,$H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$Range,$R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$The ratio of maximum height to range is given by:$\frac{H_{\max}}{R} = \frac{u^2 \sin^2 	heta / 2g}{2u^2 \sin 	heta \cos 	heta / g} = \frac{\sin^2 	heta}{4 \sin 	heta \cos 	heta} = \frac{	an 	heta}{4}$Given $	heta = 	an^{-1}(\frac{8}{7})$,so $	an 	heta = \frac{8}{7}$.Substituting the value of $	an 	heta$:$\frac{H_{\max}}{R} = \frac{8/7}{4} = \frac{8}{28} = \frac{2}{7}$.Thus,the ratio is $2:7$.

Column-$I$	Column-$II$
$i)$ The initial velocity of projection	$a)$ $\sqrt{\frac{g(1+a^2)}{2b}}$
$ii)$ The horizontal range of projectile	$b)$ $\frac{a}{b}$
$iii)$ The maximum height attained by projectile	$c)$ $\frac{a^2}{4b}$
$iv)$ The time of flight of projectile	$d)$ $a\sqrt{\frac{2}{bg}}$

$A$ body is projected from the ground at an angle of $\tan^{-1}(\frac{8}{7})$ with the horizontal. The ratio of the maximum height attained by it to its range is (in $: 7$)

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