Question

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

• A. ${\tan }^{-1}\left[\frac{g{T}^2}{2R}\right]$
• B. ${\tan }^{-1}\left[\frac{2R^2}{gT}\right]$
• C. ${\tan }^{-1}\left[\frac{2R}{g^2\pi }\right]$
• D. ${\tan }^{-1}\left[\frac{2R}{gT}\right]$

Answer 1

The correct option is A ${\tan }^{-1}\left[\frac{g{T}^2}{2R}\right]$

Time of flight, $T=\frac{2v\sin \theta }{g}$
$T^2=\frac{4v^2{\sin }^2\theta }{g^2}......\left(i\right)$
We know that,
$\text{Range},R=\frac{v^2\left(2\sin \theta \cos \theta \right)}{g}......\left(ii\right)$
Dividing $\left(i\right)$ from $\left(ii\right)$, we get
$\frac{T^2}{R}=\frac{2}{g}\tan \theta$
$\tan \theta =\frac{g{T}^2}{2R}$
$\Rightarrow \theta ={\tan }^{-1}\left[\frac{g{T}^2}{2R}\right]$