How can increased altitude affect the trajectory of a fired projectile?

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Increased altitude affects the trajectory of a fired projectile due to changes in air density and drag force. A U.S. 37-MM projectile, fired at a muzzle velocity of 792.48 m/s, requires adjustments in aiming when fired at higher altitudes, specifically a 0.28 mil or 0.01575 degree lower aim to hit a target at 800 yards. The calculations involve complex equations that account for drag force, projectile speed, and varying atmospheric conditions. The initial assumption of flat firing complicates the analysis, as the muzzle velocity changes during flight. Accurate modeling of the projectile's motion may require numerical integration of differential equations to account for these variables effectively.
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This is NOT a school work/homework, it's just a problem relate to my personal interest found on a ballastics book. And I already converted all units into SI units to make it simple.
I tried but couldn't get to correct answer.
Please not a is speed of sound, K2 is a constant with equation.
The answer is 0.28 mil lower which is also 0.01575 degree lower.

Homework Statement


A U.S. 37-MM projectile is fired with a muzzle velocity of 2600ft's (792.48m/s). The projectile weighs 1.61 Ibm(0.73kg). Assuming K2 = 0.841[unitless] and using standard sea level met data(ρ=0.0751Ibm/ft^3 (1.2kg/m^3), a = 1120 ft/s (341.4m/s))
If this weapon is used at an increased altitude and assuming the density and temperature of the atmosphere are ρ=0.06Ibm/ft^3 (0.98kg/m^3) and T=30F degree, how much higher or lower will the weapon have to be aimed to hit a target at 800 yards.

Homework Equations


Drag force: F = 1/2ρ*S*C*V^2
S = πd^2/4 (d is diameter)
Cd = ρS/2m * K2*a/Vx = k2/Vx
Vx = Vx0*exp(-k2*t) ...t is time of flight ot any range x
Vy = (Vy0+g/k2)exp(-k2*t)-g/k2
t=x/Vx0 * ln(Vx0/Vx) / (1-Vx/Vx0)
tan∅ = tan∅0 + gx/Vx0^2((1-Vx0/Vx)/(1-Vx/Vx0)) ...∠∅ is the angle of fall and ∠∅0 is the initial launch angle
y = y0 + x*tan∅0 - (g*t^2/(2ln(Vx0/Vx)))

The Attempt at a Solution



I first assume the first firing is flat firing, so the initial launch angle is zero, after the increased altitude, i calculated the new k2 and then apply it into these two equations:
Vx = Vx0*exp(-k2*t) ...t is time of flight ot any range x
Vy = (Vy0+g/k2)exp(-k2*t)-g/k2
I assume the muzzle velocity is Vx at the halfway of the range since Vy and halfway of range is zero, i then manipulate these two equations to a form of Vy0/Vx0, and the do a inverse tan to get the initial launch angle.
but it doesn't work and i realized that muzzle velocity is not Vx at halfway since the moment the projectile went out, the velocity is changing. and i was stuck here
 
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I would seem to me that in order to determine a difference in initial elevation based on different altitudes, you should be writing the differential equation describing the projectile motion and numerically integrating it. The drag force is a function of projectile speed. The drag coefficient itself is also a function of speed to some extent. There is also a slight difference in the acceleration of gravity at different altitudes if that were to be included.
 
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