A boy standing on a building of height $20\text{m}$ throws a stone with a velocity of $15{\text{ms}}^{-1}$. What should be the angle of projection so that the stone has a range of $30\text{m}$ ?
(Take $g=10m/s^2$)

The correct option is A ${\tan }^{-1}\frac{3}{2}$
Let $\theta$ be the angle of projection of the stone from the building.
Initial velocity of stone $=15{\text{ms}}^{-1}$
Initial vertical component of velocity $=15\sin \theta {\text{ms}}^{-1}$
Initial horizontal component of velocity $=15\cos \theta {\text{ms}}^{-1}$
Displacement of the stone in vertical direction when it reaches the ground $=-20\text{m}$ (Taking downward direction as positive).
Using second equation of motion, we get
$h=u_{y}t+\frac{1}{2}a{t}^2$
$\Rightarrow -20=15\sin \theta t-\frac{1}{2}g{t}^2$
$\Rightarrow -20=15\sin \theta t-5t^2$
$\begin{array}{}\Rightarrow -4=3\sin \theta t-t^2& \text{(1)}\end{array}$
We know that horizontal range
$R=u_{x}t=15\cos \theta \times t$
$\Rightarrow 30=15\cos \theta \times t$
$\begin{array}{}\Rightarrow t=\frac{2}{\cos \theta }& \text{(2)}\end{array}$
Substituting (2) in (1), we get
$-4=6\tan \theta -4se{c}^2\theta$
By using the identity $se{c}^2\theta =1+{\tan }^2\theta$
$-4=6\tan \theta -4-4{\tan }^2\theta \Rightarrow 0=4\tan \theta (\frac{3}{2}-\tan \theta )$
$\Rightarrow \tan \theta =0$ or $\tan \theta =\frac{3}{2}$
Since angle of projection can't be $0^{\circ }$,
$\theta ={\tan }^{-1}\frac{3}{2}$
Thus, the correct answer is option (a)