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

Answer 1

Answer: (A)

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