Question

A gun is mounted on a plateau $960\text{m}$ away from it’s edge as shown. Height of plateau is $960\text{m}$. The gun can fire shells with a velocity of $100\text{m/s}$ at any angle. Of the following choices, what is the minimum distance $\left(OP\right)x$ from the edge of plateau where the shell of gun can reach?

Answer 1

$P$ is point on plane for shell when it passes near edge of plateau for greater angle of projection.
In region $AP$ shell cannot hit
$x_{\text{max}}=AP$
Equation of trajectory for projectile
$y=x\tan \theta -\frac{g{x}^2}{2u^2}(1+{\tan }^2\theta )$
For point $E,y=0$ & $x=960\text{m}$
$\Rightarrow {\tan }^2\theta \times \frac{10\times (960{)}^2}{2\times (100{)}^2}-\left(\tan \theta \right)960+5\sqrt{5}=0$
$\Rightarrow \tan \theta =\frac{3}{4},\frac{4}{3}$
For $\tan \theta =\frac{4}{3},AP$ is least
Now, equation of trajectory becomes,
$y=\frac{4}{3}x-\frac{x^2}{2000}\left(\frac{25}{9}\right)$
for landing point on plane $\left(P\right),y=-960\text{m}$
$\Rightarrow -960=\frac{4}{3}x-\frac{x^2}{720}$
On solving, we get acceptable solution $x=1440\text{m}$
$AP=x-960=480\text{m}$