Question

The equation of a projectile is given by $y=\sqrt{3}x-5x^2(g=10m/s^2)$
Angle of projection is

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

Answer 1

The correct option is B ${60}^o$

$y=xtan\theta –\frac{g{x}^2}{{2v_0}^{2}co{s}^2\theta }$

So, by comparing the given equation

$y=\sqrt{3}x-5x^2$

We get that $tan\theta =\sqrt{3}$

$\theta ={60}^{\circ }$

So, Angel of projection is ${60}^{\circ }$