Question

The equation of trajectory of a projectile in a vertical planes is given by $y=ax-b{x}^2$ where $a$ and $b$ are constants and, $x$ and $y$ are the horizontal and vertical displacements of the particle from the point of projection. The angle of projection of projectile from the horizontal is

• A. ${\tan }^{-1}\left[\frac{1}{b}\right]$
• B. ${\tan }^{-1}\left[\frac{a}{b}\right]$
• C. ${\tan }^{-1}\left[2\alpha \right]$
• D. ${\tan }^{-1}\left[a\right]$

Answer 1

The correct option is D

${\tan }^{-1}\left[a\right]$

Step 1: Given data

The equation of trajectory is $y=ax-b{x}^2$

Step 2: Find the angle of projection of the projectile from the horizontal

Compare the equation of trajectory of the projectile motion of a particle with the given equation of trajectory

$$\begin{gathered} \Rightarrow ax-b{x}^2=x\tan \theta -\frac{g}{2u^2{\cos }^2\theta }{x}^2 \\ \Rightarrow a=\tan \theta \\ \Rightarrow \theta ={\tan }^{-1}a \end{gathered}$$

Hence, option D is correct.