Question

If the time of flight of a bullet over a horizontal range $R$ is $T$, then the angle of projection with horizontal is

• A. $\tan ^{-1}\left(\frac{g T^{2}}{2 R}\right)$
• B. $\tan ^{-1}\left(\frac{2 R^{2}}{g T}\right)$
• C. $\tan ^{-1}\left(\frac{2 R}{g^{2} T}\right)$
• D. $\tan ^{-1}\left(\frac{2 R}{g T}\right)$

Answer 1

Answer: (A)

$T=\frac{2 u \sin \theta}{g}$
$\Rightarrow u=\frac{g T}{2 \sin \theta}$
$R=\frac{2 u^{2} \sin \theta \cos \theta}{g}$
$R=\frac{2 u \sin \theta}{g} \times u \cos \theta$
$R=T \times u \cos \theta$
$R=T \times \frac{g T \cos \theta}{2 \sin \theta}$
$R=\frac{g T^{2}}{2} \frac{1}{\tan \theta}$
$\tan \theta=\frac{g T^{2}}{2 R}$
$\theta=\tan ^{-1}\left(\frac{g T^{2}}{2 R}\right)$