Question

The equation of a projectile is $y=\sqrt{3}$ $x-\frac{g{x}^2}{2}$ the angle of projection is :

• A. ${30}^0$
• B. ${60}^0$
• C. ${45}^0$
• D. None

Answer 1

The correct option is B ${60}^0$

the correct answer is B.

the equation of the projectile is

$Y=\sqrt{3}x-\frac{g{x}^2}{2}$-------(1)

we know that if a projectile is projected with speed $u$ at angle $\theta$ with horizontal.

then, the equation of trajectory of a projectile is:

$Y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$--------(2)

comparing both equation(1)&(2)

$\sqrt{3}x-\frac{g{x}^2}{2}=tan\theta x-\frac{g{x}^2}{2u^{2}co{s}^2\theta }$

$tan\theta =\sqrt{3}=tan{60}^0$

So, $\theta ={60}^0$