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

Let stone is projected at an angle $\alpha$ with horizontal.

$\therefore (u\cos \alpha )t=\text{ Range }=R$

$\text{ and }-3=(10\sin \alpha )t-\frac{1}{2}\times 10\times t^2$

From Eqs. (i) and (ii), we get

$\begin{array}{l}{R}^2+{\left(5t^2-3\right)}^2=100t^2\\ \Rightarrow R^2+25t^4+9-30t^2=100t^2\\ \Rightarrow 25t^4-130t^2+9+R^2=0\end{array}$

The discriminant of quadratic in t² should be greater or equal to zero.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}D\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-130{)}^2-4\times 25\left(9+R^2\right)\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}16900\ge 100\left(9+R^2\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(9+R^2\right)\le 169\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}9+R^2\le 169\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{R}^2\le 160\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}R\le \sqrt{160}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{R}_{max}=4\sqrt{10}\mathrm{m}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=4\end{array}$