Question

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

Answer 1

Answer: 4

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