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 150ms^-1.

Dear Student,
Here in this question the given data ,
Height of the tower = h= 100 m,
Muzzle speed of bullet = u = 150 m/s
Let $\theta$ be the angle at which machine gun would fire in order to cover maximum distance .
Then the horizontal component of velocity = $150 \cos \theta$
and the vertical component of velocity = $150 \sin \theta$
If T is the time of flight then ,horizontal range ,R = $\left(150 \cos \theta \right)\times T$
Now as the gun is mounted on the tower of 100 m high .
Therefore, the positive direction of the position axis as to be along the line from the top of the tower in downward direction .
For motion along vertical :
Initial velocity = $-150 \sin \theta$
Distance covered = 100 m
Acceleration due to gravity = g= 10 m/s²
Now in time T the machine gun shot will reach maximum height and then reach the ground .
Now from laws of motion , 
$$\begin{gathered} S= ut +\frac{1}{2}a{t}^2 \\ \Rightarrow 100=\left(-150\sin \theta \right)\times T +\frac{1}{2}\times 10\times T^2 \\ \Rightarrow T^2-\left(30 \sin \theta \right)\times T -20 =0 \\ \Rightarrow T=\frac{30\sin \theta \pm \frac{1}{2}\left(900\sin 2\theta +80\right)}{2} \\ \Rightarrow T =15\sin \theta \pm \left(225\sin 2\theta +20\right) \end{gathered}$$
Now the range will be maximum if the flight time is maximum .
Therefore positive sign we have ,
$T =15\sin \theta +\left(225\sin 2\theta +20\right)$
Hence horizontal range covered,
^{$R=150\cos \theta \times \left[15\sin \theta +\left(225\sin 2\theta +20\right)\right]$}
Now the horizontal range is maximum when $\theta =45°$
But in this case the machine gun is mounted at the height of 100 m
So at $\theta =45°$ the range will not be maximum , it will be maximum for some value of $\theta$ which are close to 45⁰ .
Now if we calculate the values of R by setting $\theta =43°,43·5°,44°,45°,46°,47°$ the values of R come out to be $2347 m, 2347·7m,2348m, 2346m,2341m,2334m$
Thus R is maximum for some value $\theta$ between $43° and 43·5°$ .
Therefore the mean value of $\theta =\frac{43+43·5}{2}=43·75°$ .
Therefore the gun should be inclined at $\theta =43·75°$ to cover maximum range firing on the ground below .
Regards