Question

The equation of a projectile is $ y = \sqrt 3 x - \dfrac{{g{x^2}}}{2} $ . The angle of projection is:
(A) $ {30^o} $
(B) $ {60^o} $
(C) $ {45^o} $
(D) None

Answer 1

Hint: We are given with the equation of the projectile and are asked to find the angle with which it was projected. Thus, we will compare the given formula of projectile to the generic formula for the same. And then calculate the angle of projection from a suitable parameter.

Formulae Used
 $ y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }} $
Where, $ \theta $ is the angle of projection and $ u $ is the initial velocity with which the body was projected.

Complete step by step answer:
Here, We are given that,
 $ y = \sqrt 3 x - \dfrac{{g{x^2}}}{2} \cdot \cdot \cdot \cdot (1) $
But, We know,
 $ y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }} \cdot \cdot \cdot \cdot (2) $
Comparing equations $ (1) $ and $ (2) $ , we can say
 $ \tan \theta = \sqrt 3 $
Thus, We get,
 $ \theta = {60^0} $
Hence, the correct option is (B).

Additional Information:
A projectile can be defined as any object thrown or launched by applying a certain force on it. Thus, after the application of force, the object or the projectile starts moving in a certain direction as per Newton's laws of motion. The path the projectile follows before coming to a halt is termed as the trajectory of the projectile. The shape and nature of the trajectory can be theoretically evaluated by the equation which governs the motion of the projectile.
After having said that, there will certainly be some fundamental parameter which will assist the nature and shape of the trajectory or the motion of the projectile.
These parameters are basically as listed down below:
-Range of the projectile: This is the parameter which tells us the displacement of the object in the horizontal direction from the point of launch.
-Height of the projectile: This parameter tells us about the maximum height attained by the projectile while its entire motion.
-Time Period: This parameter tells us the time taken by the projectile to complete the entire trajectory.

Note:
Here, the angle with which the projectile was launched was $ {\tan ^{ - 1}}\sqrt 3 $ which can attain values of $ {60^o},{240^o},... $ . But, we have taken the value in the principal branch of this function. Thus, the value turns out to be $ {60^o} $ .