The maximum height reached by the projectile is $4m$. The horizontal range is $12m$. Velocity of projection (in $m{s}^{-1}$) is
($g$ is acceleration due to gravity)

The correct option is A $5\sqrt{\frac{g}{2}}$
In a projectile projected at angle $\theta$ maximum height and range is given by
$H_{max}=\frac{u^2{\sin }^2\theta }{2g}=4.....\text{(i)}{R}_{max}=\frac{u^2\sin 2\theta }{g}=12......\text{(ii)}$
Dividing equation $\text{(ii)}$ by equation $\text{(i)}$
$3=\frac{\sin 2\theta }{{\sin }^2\theta }\times 2\frac{3}{2}=\frac{2\sin \theta \cos \theta }{{\sin }^2\theta }\tan \theta =\frac{4}{3}$
$\Rightarrow \sin \theta =\frac{4}{5}\text{Using equation}\text{(i)}4=\frac{u^2}{2g}\times {\left(\frac{4}{5}\right)}^2{u}^2=\frac{25g}{2}u=5\sqrt{\frac{g}{2}}m/s$