If R and H are the horizontal range and maxim | vedclass

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(B) The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$,which implies $\sin 	heta = \sqrt{\frac{2gH}{u^2}}$.The horizontal range $R

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$\sqrt{2 g H+\frac{R^2 g}{8 H}}$

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(B) The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$,which implies $\sin 	heta = \sqrt{\frac{2gH}{u^2}}$.The horizontal range $R$ is given by $R = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.Substituting $\sin 	heta$ and $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{2gH}{u^2}}$ into the range formula:$R = \frac{2u^2}{g} \cdot \sqrt{\frac{2gH}{u^2}} \cdot \sqrt{1 - \frac{2gH}{u^2}} = \frac{2u^2}{g} \cdot \frac{\sqrt{2gH}}{u} \cdot \sqrt{\frac{u^2 - 2gH}{u^2}}$.Simplifying this,we get $R = \frac{2}{g} \cdot \sqrt{2gH} \cdot \sqrt{u^2 - 2gH}$.Squaring both sides:$R^2 = \frac{4}{g^2} \cdot 2gH \cdot (u^2 - 2gH) = \frac{8H}{g} (u^2 - 2gH)$.Rearranging for $u^2$:$u^2 - 2gH = \frac{R^2 g}{8H} \Rightarrow u^2 = 2gH + \frac{R^2 g}{8H}$.Thus,the speed of projection is $u = \sqrt{2gH + \frac{R^2 g}{8H}}$.

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

If $R$ and $H$ are the horizontal range and maximum height attained by a projectile,then its speed of projection is ..........

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