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{gT^{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,
$T=\frac{2 u \sin 45^{\circ}}{g}$
$=\frac{2 u}{\sqrt{2} \cdot g}=\frac{\sqrt{2} \cdot u}{g}$
Or $u=\frac{T g}{\sqrt{2}}$
Putting these value of $u$ in Eq. (i), we get
$R_{\max }=\frac{1}{g}\left(\frac{T g}{\sqrt{2}}\right)^{2} $
$R_{\max }=\frac{1}{2} g\, T^{2}$