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 θ be the angle at which machine gun would fire in order to cover maximum distance .
Then the horizontal component of velocity = 150 cosθ
and the vertical component of velocity = 150 sinθ
If T is the time of flight then ,horizontal range ,R = 150 cosθ×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 sinθ
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 , 
S= ut +12at2100=-150sinθ×T +12×10×T2T2-30 sinθ×T -20 =0T=30sinθ±12900sin2θ+802T =15sinθ±225sin2θ+20
Now the range will be maximum if the flight time is maximum .
Therefore positive sign we have ,
T =15sinθ+225sin2θ+20
Hence horizontal range covered,
Now the horizontal range is maximum when θ=45°
But in this case the machine gun is mounted at the height of 100 m
So at 
θ=45° the range will not be maximum , it will be maximum for some value of θ which are close to 450 .
Now if we calculate the values of R by setting θ=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 θ between 43° and 43·5° .
Therefore the mean value of θ=43+43·52=43·75° .
Therefore the gun should be inclined at θ=43·75° to cover maximum range firing on the ground below .


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