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(C) The range of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.For maximum range,the angle of projection is $	heta = 45^{\circ}$.Thus,$R_

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$\frac{g T^2}{2}$

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(C) The range of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.For maximum range,the angle of projection is $	heta = 45^{\circ}$.Thus,$R_{\max} = \frac{u^2 \sin(2 	imes 45^{\circ})}{g} = \frac{u^2}{g} \quad \dots(i)$.The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.Substituting $	heta = 45^{\circ}$,we get $T = \frac{2u \sin 45^{\circ}}{g} = \frac{2u}{g \sqrt{2}} = \frac{u \sqrt{2}}{g}$.From this,we find $u = \frac{Tg}{\sqrt{2}}$.Substituting the value of $u$ into equation $(i)$:$R_{\max} = \frac{1}{g} \left( \frac{Tg}{\sqrt{2}} ight)^2 = \frac{1}{g} \left( \frac{T^2 g^2}{2} ight) = \frac{g T^2}{2}$.

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight,then the value of maximum range is (acceleration due to gravity $= g$)

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