The maximum height attained by a projectile when thrown at an angle $\theta $ with the horizontal is found to be half the horizontal range. Then, $\theta $ is equal to:
$\text{A}\text{. }{{\tan }^{-1}}(2)$
$\text{B}\text{. }\dfrac{\pi }{6}$
$\text{C}\text{. }\dfrac{\pi }{4}$
$\text{D}\text{. }{{\tan }^{-1}}\left( \dfrac{1}{2} \right)$

Hint: Use the horizontal and vertical components of the velocity by which the projectile is fired. Find the expressions for maximum height and horizontal range in terms of common quantities. Then you can equate the expressions as per the given relation between the maximum height and horizontal range.

Complete step by step answer:
It is given that the angle at which the projectile is thrown is $\theta $. Let its initial velocity (velocity at which the projectile is thrown) be u. Since the projectile is under the influence of gravity, the acceleration (a) of the projectile will be g downwards.
Let the maximum height attained by the projectile be H and the horizontal range of the projectile be R.
It is given that $H=\dfrac{R}{2}$ …………(i).
First we will find the expressions for H and R in terms of u and $\theta $ and then equate according to equation (i).
Resolve the velocity vector u into horizontal (${{u}_{x}}$) and vertical (${{u}_{y}}$) components.
${{u}_{x}}=u\cos \theta $.
${{u}_{y}}=u\sin \theta $
When the projectile is at its maximum height, its displacement (s) is equal to H and also its velocity at this point is equal to zero.
Here, we can use the kinematic equation, $2as={{v}^{2}}-{{u}^{2}}$ for the of u in vertical direction and substitute and the values of a, s, u and v as g, H, ${{u}_{y}}=u\sin \theta $ and 0 respectively.
Therefore we get,
$2(-g)(H)={{0}^{2}}-{{\left( u\sin \theta \right)}^{2}}$
$\Rightarrow -2gH=-{{u}^{2}}{{\sin }^{2}}\theta $
Therefore, $H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ ……(ii)
For R we will use the horizontal component of u i.e. ${{u}_{x}}=u\cos \theta $.
Here, since there is no force that is affecting the horizontal velocity, its acceleration in this direction is zero.
Range is the horizontal displacement of the projectile when it hits the ground.
Therefore, we can simply use the formula, $\text{displacement= velocity }\!\!\times\!\!\text{ time}$.
But, for this we need the time taken by the projectile to hit the ground.
Let that time be T.
We know that when the ball hits the ground, the vertical displacement of the projectile is zero.
To find T, we can use the kinematic equation, $s=ut+\dfrac{1}{2}a{{t}^{2}}$ ………(iii).
Here, s=0, $u={{u}_{y}}=u\sin \theta $, t=T and a = -g.
Substitute the respective values in equation (iii).
Therefore, we get,
$0=u\sin \theta T+\dfrac{1}{2}(-g){{T}^{2}}$
$\Rightarrow \dfrac{1}{2}g{{T}^{2}}=u\sin \theta T$
Divide both sides by T.
$\Rightarrow \dfrac{1}{2}gT=u\sin \theta $.
$\Rightarrow T=\dfrac{2u\sin \theta }{g}$
And we know that $R={{u}_{x}}T=u\cos \theta T$
Therefore, $R=u\cos \theta .\left( \dfrac{2u\sin \theta }{g} \right)$
$\Rightarrow R=\dfrac{2{{u}^{2}}\sin \theta \cos \theta }{g}$ ……… (iv)
From (i) , (ii) and (iii) we get,
$\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\left( \dfrac{\dfrac{2{{u}^{2}}\sin \theta \cos \theta }{g}}{2} \right)$
$\Rightarrow {{\sin }^{2}}\theta =2\sin \theta \cos \theta $
Subtracting $2\sin \theta \cos \theta $ on both the side, we get
$\Rightarrow {{\sin }^{2}}\theta -2\sin \theta \cos \theta=0 $
taking $\sin \theta$ common, we will get
$\Rightarrow {\sin \theta({\sin \theta -2\cos \theta})}=0 $
$\Rightarrow \sin \theta=0$ which is not possible or
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }=2$
$\Rightarrow \tan \theta =2\Rightarrow \theta ={{\tan }^{-1}}(2)$.
Hence, the correct option is A.

Note: Try to remember the formulas of maximum height, time of flight and horizontal range of the projectile. It comes handy when we solve questions of this type. In this solution, we derived these formulas. It is not compulsory to derive these in every question. You can use them directly whenever required.