Two balls are projected from the same point directions at $30^\circ $ and $60^\circ $ with the horizontal. Both the balls attain the same height. The ratio of their velocity of projection is
A. $\sqrt 3 :2$
B. $\sqrt 3 :3$
C. $\sqrt 3 :5$
D. $\sqrt 3 :1$

Hint: In order to solve this question we will understand the definition of projectile motion which states that projectile motion is that motion in which the trajectory of a particle is parabolic in nature. It is motion that is experienced by a launched object. It can occur above earth or at surface where various factors like initial speed of motion, direction with horizontal and the direction of gravity decides range up-to which particle is thrown and height it can achieve.

Formula used:
Total time of flight of a projectile motion is calculated as,
$T = \dfrac{{(2u\sin \theta )}}{g}$
Maximum height attained by projectile motion is calculated as,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
where, $u$ is the initial velocity of projectile motion, $\theta $ is the angle made by the projectile with the horizontal and $g$ is the acceleration due to gravity.

Complete step by step answer:
Let us first calculate height which could be achieved by a particle with speed $u$ and let the angle it makes horizontal is $\theta $. Also the value of acceleration due to gravity is $ - g$ where $g = 9.8\,m\,{s^{ - 2}}$.
Since we know the time of flight given by $T = \dfrac{{(2u\sin \theta )}}{g}$.
So the time taken to reach max height is $t = \dfrac{T}{2} = \dfrac{{u\sin \theta }}{g}$.
So the maximum height that could be achieved is given by
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$

Since max height attained by balls is same so let speed of first ball be ${u_1}$ and angle be $\theta = 30^\circ $
Similarly for second ball let the speed by ${u_2}$ and angle be $\theta = 60^\circ $
So using relation for same height we get
${H_1} = {H_2}$
\[\Rightarrow \dfrac{{{u_1}^2{{\sin }^2}30}}{{2g}} = \dfrac{{{u_2}^2{{\sin }^2}60}}{{2g}}\]
$\Rightarrow \dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{{{{\sin }^2}60}}{{{{\sin }^2}30}}$
$\Rightarrow \dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{{(\dfrac{3}{4})}}{{(\dfrac{1}{4})}} \\
\Rightarrow \dfrac{{{u_1}^2}}{{{u_2}^2}}= \dfrac{3}{1}$
$ \therefore \dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{\sqrt 3 }}{1}$

So correct option is D.

Note: It should be remembered that here we have assumed the balls are thrown from the earth surface but in some cases if the ball is thrown from a window or from a building only the value of acceleration changes but the final result is the same. Also time taken by particles to reach maximum height is half of total time flight because of the symmetrical path of projectile motion.