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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 the bullet is \[150m/s\]. Take \[g = 10m{s^{ - 2}}\]

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Last updated date: 21st Apr 2024
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Answer
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Hint: The motion of the bullet is a projectile motion. Find the locus of the projectile and simplify to get a clear relation between the angle and the height of the tower and the velocity of the bullet. Put the values that are given in the problem and find the angle.

Formula used:
\[y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\]
\[y = \] vertical position = the height of the tower.
\[x = \] horizontal position = the maximum range of firing.
\[u = \] Initial velocity or the muzzle speed of the bullet.
\[g = \] acceleration due to gravity.
\[\theta = \] the angle of the initial velocity from the horizontal plane (radians or degrees)

Complete step by step answer:
Let us consider a frame of reference where the positive y-axis is extended vertically and the positive x-axis is along with the projectile velocity horizontally. The main point is the projection point. The locus of the projectile is given by,
\[y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}...............(1)\]
\[y = \] vertical position = the height of the tower = $ - h$ (negative sign implies the opposite direction)
\[x = \] horizontal position = the maximum range of firing = $R$
\[u = \] initial velocity or the muzzle speed of the bullet
\[g = \] acceleration due to gravity
\[\theta = \] the angle of the initial velocity from the horizontal plane (radians or degrees)
So, eq. (1) can be written as
\[\Rightarrow - h = R\tan \theta - \dfrac{{g{R^2}}}{{2{u^2}{{\cos }^2}\theta }}\]
\[ \Rightarrow h = - R\tan \theta + \dfrac{{g{R^2}}}{{2{u^2}}}{\sec ^2}\theta \]
Differentiating w.r.t \[\theta \] ,
\[\Rightarrow 0 = - R{\sec ^2}\theta - \tan \theta \dfrac{{dR}}{{d\theta }} + \dfrac{{gR}}{{2{u^2}}}{\sec ^2}\theta + g\dfrac{{{{\sec }^2}\theta }}{{2{u^2}}}2R\dfrac{{dR}}{{d\theta }}..........(2)\]
Since we need the maximum range,
\[\Rightarrow \dfrac{{dR}}{{d\theta }} = 0\]
\[ \Rightarrow R = \dfrac{{{u^2}}}{{g\tan \theta }}\]
Putting these values in (2) we get,
\[\Rightarrow h = \dfrac{{ - {u^2}}}{g} + \dfrac{{g{{\sec }^2}\theta }}{{2{u^2}}} \times \dfrac{{{u^4}}}{{{g^2}{{\tan }^2}\theta }}\]
\[\Rightarrow \sin \theta = \sqrt {\dfrac{{{u^2}}}{{2({u^2} + gh)}}} ..................(3)\]
Given, $u = 150\,m/s$
$\Rightarrow g = 10\,m/{s^2}$
$\Rightarrow h = 100\,m$
\[\Rightarrow \sqrt {\dfrac{{{u^2}}}{{2({u^2} + gh)}}} = \sqrt {\dfrac{{{{150}^2}}}{{2({{150}^2} + 10 \times 100)}}} \]
$ \Rightarrow 0.691$
Put this value in equation (3) we get,
\[\Rightarrow \sin \theta = 0.691\]
$ \therefore \theta = 43.70$

So, the gun is inclined to cover a maximum range of firing on the ground below at an angle $\theta = 43.70^\circ $.
Note: The locus of the projectile is given by,
\[y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\]
This equation is in terms of $y = a x + b{x^2}$. This equation is the equation of the locus of a parabola. So we must include that the locus of a projectile is Parabolic.