A stone is projected from top of a vertical pole of height 3 m with initial velocity $10m{s}^{-1}.$ The maximum range on the ground is $\sqrt{10}xm.$ Find the value of x. 
              
 

Let stone is projected at an angle $\alpha$ with horizontal.
$\therefore \left(u\cos \alpha \right)t=Range=R$ ………. (i)
And $-3=\left(10\sin \alpha \right)t-\frac{1}{2}\times 10\times t^2$ ……..(ii)
From Eqs. (i) and (ii), we get 
$R^2+{(5t^2-3)}^2=100t^2$ 
$\Rightarrow R^2+25t^4+9-30t^2=100t^2$
$\Rightarrow 25t^4-130t^2+9+R^2=0$
The discriminant of quadratic in $t^2$ should be greater or equal to zero.
$\therefore D\ge 0$
$\Rightarrow {(-130)}^2-4\times 25(9+R^2)\ge 0$
$\Rightarrow 16900\ge 100(9+R^2)$
$\Rightarrow (9+R^2)\le 169\Rightarrow 9+R^2\le 169\Rightarrow R^2\le 160$
$\therefore R\le \sqrt{160}$
$\therefore R_{\max }=4\sqrt{10}m$
$\therefore x=4$