Question

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 $\sqrt{10} x \,m$. The value of $x$ is ______

Answer 1

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(5 t^{2}-3\right)^{2}=100 t^{2} $
$\Rightarrow R^{2}+25 t^{4}+9-30 t^{2}=100 t^{2} $
$\Rightarrow 25 t^{4}-130 t^{2}+9+R^{2}=0$
The discriminant of quadratic in $t^{2}$ should be greater or equal to zero.
$\therefore D \geq 0$
$\Rightarrow (-130)^{2}-4 \times 25\left(9+R^{2}\right) \geq 0$
$\Rightarrow 16900 \geq 100\left(9+R^{2}\right) $
$\Rightarrow \left(9+R^{2}\right) \leq 169$
$\Rightarrow 9+R^{2} \leq 169$
$\Rightarrow R^{2} \leq 160 $
$\therefore R \leq \sqrt{160}$
$\therefore R_{\max }=4 \sqrt{10} \,m$
$\therefore x=4$