Question

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

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

Answer 1

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


$T=\frac{2vsin\theta }{g^2}$

$T^2=\frac{4v^{2}si{n}^2\theta }{g^2}$

$R=\frac{v^2\left(2sin\theta cos\theta \right)}{g}$

$\frac{T^2}{R}=\frac{2}{g}tan\theta$

Or $tan\theta =\frac{g{T}^2}{2R}$

Or $\theta =ta{n}^{-1}\left[\frac{g{T}^2}{2R}\right]$