# 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 =
and the vertical component of velocity =
If T is the time of flight then ,horizontal range ,R =
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 =
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 ,

Now the range will be maximum if the flight time is maximum .
Therefore positive sign we have ,

Hence horizontal range covered,
$R=150\mathrm{cos}\theta ×\left[15\mathrm{sin}\theta +\left(225\mathrm{sin}2\theta +20\right)\right]$
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
Thus R is maximum for some value $\theta$ between  .
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

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