Question

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 $=g$ )

• A. $\frac{g^{2}T}{2}$
• B. $\frac{gT}{2}$
• C. $\frac{g{T}^2}{2}$
• D. $\frac{g^2{T}^2}{2}$

Answer 1

Answer: (C)

We know that,
Range of projectile
$R=\frac{u^2\sin 2\theta }{g}$
As range is maximum, $\theta =45^{\circ }$
$R_{\max }=\frac{u^2\sin 2\times 45}{g}=\frac{u^2}{g}...(i)$
Flight time of projectile,
$T=\frac{2u\sin 45^{\circ }}{g}$
$=\frac{2u}{\sqrt{2}\cdot g}=\frac{\sqrt{2}\cdot u}{g}$
Or $u=\frac{Tg}{\sqrt{2}}$
Putting these value of $u$ in Eq. (i), we get
$R_{\max }=\frac{1}{g}{\left(\frac{Tg}{\sqrt{2}}\right)}^2$
$R_{\max }=\frac{1}{2}g{T}^2$