Cannon Shot With Air Resistance

If we were to ignore air resistance, we would aim the gun at an angle $\theta$ , where $\frac{v_0^2\sin2\theta}{g} = 2000$ , and hence $\sin2\theta = \tfrac29$ . Since $\theta \ge 40^\circ$ , we deduce that $\theta = 83.5802^\circ$ .

With air resistance, the equation of motion of the cannon-ball is $\begin{aligned} \tfrac43\pi r^3 \rho_{\text{steel}} \ddot{\mathbf{r}} & = -\tfrac12\rho_{\text{air}} \pi r^2 |\dot{\mathbf{r}}| \dot{\mathbf{r}} - \tfrac43\pi r^3 \rho_{\text{steel}} g \binom{0}{1} \\ \ddot{\mathbf{r}} & = \; -\frac{3\rho_{\text{air}}}{8\rho_{\text{steel}} r}|\dot{\mathbf{r}}| \dot{\mathbf{r}} - g\binom{0}{1} \end{aligned}$ If we solve this equation numerically (I used $dt = 0.0001$ for sufficient accuracy) with $\mathbf{r}(0) \; = \; \binom{0}{0} \hspace{2cm} \dot{\mathbf{r}}(0) \; = \; 300\binom{\cos83.5802^\circ}{\sin83.5802^\circ}$ we deduce that the cannon-ball hits the ground at the point $\binom{410.97}{0}$ which represents a range short-fall of $\boxed{1589.03}$ meters.

It is worth noting that the angle of projection needed to reach the target, taking air resistance into account, is roughly $\theta = 42.06^\circ$ .

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while y >= 0:
    ax = - C_d * (3 * rho_air * math.sqrt(vx ** 2 + vy ** 2) * vx)/(8 * rho_steel * r)
    ay = - C_d * (3 * rho_air * math.sqrt(vx ** 2 + vy ** 2) * vy)/(8 * rho_steel * r) - g
    (x, y) = (x + vx * dt + 0.5 * ax * dt ** 2, y + vy * dt + 0.5 * ay * dt ** 2)
    (vx, vy) = (vx + ax * dt, vy + ay * dt)

I do object to the Ottoman invasion and the siege of Constantinople. If I could change history, I'd let the yahoos have their way and laugh at their lopping of projectiles on their own siegeworks. Or, I would take my place on the city wall and shoot at an angle of about $42^\circ$ into the Ottoman invaders, which would give the proper range of 2000 m.

Note that in my simulation I use $\Delta \vec x = \tfrac12(\vec v_{\text{old}} + \vec v_{\text{new}})\:dt.$ It is a very simple way to obtain more accuracy.

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import math

rho_air = 1.22
g = 10.0
rho_steel = 8050.0
v0 = 300.0
r = 0.08
ran = 2000
drag = "On" # Set this to "On"  or "Off" to include drag, or not.
C_d = (1 if drag is "On" else 0)

theta = 90 - math.asin(ran*g/(v0*v0))/2 * 180/math.pi

print("The launch angle is {:.4f} degrees.".format(theta))

(vx, vy) = (v0 * math.cos(theta * math.pi / 180), v0 * math.sin(theta * math.pi / 180))
(x, y) = (0, 0)

dt = 0.001
A = math.pi * r*r
m = 4*math.pi/3 * r*r*r * rho_steel

while y >= 0:
    v2 = vx*vx + vy*vy
    Fd = -0.5*rho_air*C_d*A*v2
    v = math.sqrt(v2)
    (Fdx, Fdy) = (Fd*vx/v, Fd*vy/v)
    (ax, ay) = (Fdx/m, Fdy/m - g)
    (ovx, ovy) = (vx, vy)
    (vx, vy) = (vx + ax*dt, vy + ay*dt)
    (x, y) = (x + (vx+ovx)/2*dt, y + (vy+ovy)/2*dt)

print("The cannonball's range is {:.2f} meters.".format(x))

Output:

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The launch angle is 83.5802 degrees.
The cannonball's range is 410.93 meters.