Question

The parabolic path of a projectile is represented by $y=\frac{x}{\sqrt{3}}-\frac{x^2}{60}$ in MKS units. It's angle of projection is ($g=10m{s}^{-2}$)

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is A ${30}^o$

$y=\frac{x}{\sqrt{3}}-\frac{x^2}{60}$
Angle of projection is given by , $\frac{dy}{dx}{|}_{x=0}=\tan \theta$

$\frac{dy}{dx}=\frac{1}{\sqrt{3}}-\frac{2x}{60}{|}_{x=0}\Rightarrow \frac{1}{\sqrt{3}}=\tan \theta$

$\Rightarrow \theta ={30}^0$