Question

The parabolic path of a projectile is represented by $y = \dfrac{x}{{\sqrt 3 }} - \dfrac{{{x^2}}}{{60}}$ in MKS units. Its angle of projection is $\left( {g = 10{\text{m}}{{\text{s}}^{ - 1}}} \right)$
(A) ${30^o}$
(B) ${45^o}$
(C) ${60^o}$
(D) ${90^o}$

Answer 1

Hint: The general equation of trajectory of projectile projected with initial velocity $u$ and with angle $\theta $ with the horizontal is given by
 $y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$
On converting the equation given in the question to the form of the standard equation of trajectory of the projectile and then comparing the coefficient of x and y we can easily determine the value of the angle of projection with which the projectile is projected.

Complete step by step solution:
According to the initial data we have
The given equation is $y = \dfrac{x}{{\sqrt 3 }} - \dfrac{{{x^2}}}{{60}}$ in MKS units.
The general equation of trajectory of projectile projected with initial velocity $u$ and with angle $\theta $ with the horizontal is given by
$y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}{\text{ }}...{\text{(1)}}$
On comparing the given equation $y = \dfrac{x}{{\sqrt 3 }} - \dfrac{{{x^2}}}{{60}}$ with equation (1) .
 Now, on comparing the coefficient of x we have,
$\tan \theta = \dfrac{1}{{\sqrt 3 }}$
$ \Rightarrow \theta = {30^o}$
Hence the angle of projection with which the projectile is projected is ${30^o}$.

Hence, the option (a) is correct.

Note: The trajectory formula helps us to find the gravity that acted on an object. Also, the trajectory has vertical (y) and horizontal (x) position components. Moreover, if we launch the projectile with an initial velocity v at an angle θ from the horizontal plane. Then we can find the vertical position of the object from the horizontal position of the object also by the equation of trajectory of the projectile.