The equation of a projectile is $$y = \sqrt 3 x - \frac{{g{x^2}}}{2}$$
The angle of projection is given by
A.
$$\tan \theta = \frac{1}{{\sqrt 3 }}$$
B.
$$\tan \theta = \sqrt 3 $$
C.
$$\frac{\pi }{2}$$
D.
zero.
Answer :
$$\tan \theta = \sqrt 3 $$
Solution :
Comparing the given equation with
$$y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }},$$
we get $$\tan \theta = \sqrt 3 $$
Releted MCQ Question on Basic Physics >> Kinematics
Releted Question 1
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