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(C) The standard equation of a projectile is $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with $y = ax - bx^2$,we get $	an

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$i-d, ii-a, iii-c, iv-b$

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(C) The standard equation of a projectile is $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with $y = ax - bx^2$,we get $	an 	heta = a$ and $b = \frac{g}{2u^2 \cos^2 	heta}$.$i)$ Initial velocity $u$: Since $	an 	heta = a$,$\sec^2 	heta = 1 + a^2$,so $\cos^2 	heta = \frac{1}{1+a^2}$. Substituting into $b$,$b = \frac{g(1+a^2)}{2u^2} \implies u = \sqrt{\frac{g(1+a^2)}{2b}}$. Thus,$i-d$.$ii)$ Range $R$: $y=0 \implies x(a-bx)=0 \implies R = \frac{a}{b}$. Thus,$ii-a$.$iii)$ Max height $H$: $H = \frac{u^2 \sin^2 	heta}{2g} = \frac{u^2 	an^2 	heta \cos^2 	heta}{2g} = \frac{a^2}{4b}$. Thus,$iii-c$.$iv)$ Time of flight $T$: $T = \frac{2u \sin 	heta}{g} = \frac{2u 	an 	heta \cos 	heta}{g} = a \sqrt{\frac{2}{bg}}$. Thus,$iv-b$.Therefore,the correct match is $i-d, ii-a, iii-c, iv-b$.

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	$d)$ $a\sqrt{\frac{2}{bg}}$

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