Question

A machine gun is mounted on the top of a tower $100m$ high. At what angle should the gun be inclined to cover a maximum range of firing on the ground below? The muzzle speed of bullet is $150m{s}^{-1}$. Take $g=10m{s}^{-2}$.

Answer 1

$-h=R\tan \theta -\frac{g{R}^2}{2u^2{\cos }^2\theta }$
$\Rightarrow h=-R\tan \theta +\frac{g{R}^2}{2u^2}{\sec }^2\theta .....\left(i\right)$
Differentiating w.r.t $\theta$
$\Rightarrow 0=-R{\sec }^2\theta -\tan \theta \frac{dR}{d\theta }+\frac{gR}{2u^2}2{\sec }^2\theta +\frac{g{\sec }^2\theta }{2u^2}2R\frac{dR}{d\theta }$
Putting $\frac{dR}{d\theta }=0$, we get $R=\frac{u^2}{g\tan \theta }$. Putting $R$ in (i), we get
$h=-\frac{u^2}{g}+\frac{g{\sec }^2\theta }{2u^2}\frac{u^4}{g^2{\tan }^2\theta }$
$\Rightarrow \sin \theta =\sqrt{\frac{u^2}{2(u^2+gh)}}\Rightarrow \theta ={43.78}^{\circ }$.