Case Study 1 — The Artillery Officer's Problem: Range, Angle, and the Parabola

Field: Physics and ballistics Calculus used: Parametric trajectories (§25.6), the chain-rule slope (§25.3), optimization (Chapter 10), eliminating the parameter (§25.2)

A shell, a hill, and a single number

In the spring of 1916, somewhere behind the lines, a young artillery officer is handed a problem that calculus had solved two centuries earlier but that his gunnery tables solve only crudely. A target sits 6,000 meters away. His gun fires a shell at a muzzle speed he can measure but not change much — call it $v_0 = 250$ m/s. The one knob he controls is the elevation: the angle $\theta$ at which the barrel points above the horizontal. Set it too low and the shell falls short; too high and it sails over. He needs the angle that lands the shell exactly on the target, and he needs to know whether the target is even reachable at all.

Strip away the uniform and this is precisely the parametric model of §25.6. The shell leaves the muzzle (which we place at the origin) and, ignoring air resistance for now, traces

$$x(t) = (v_0\cos\theta)\,t, \qquad y(t) = (v_0\sin\theta)\,t - \tfrac{1}{2}g t^2,$$

with $g = 9.8\ \text{m/s}^2$. The parameter $t$ is genuinely time. The horizontal equation carries one physical fact — no horizontal force, so constant speed $v_0\cos\theta$ — and the vertical equation carries another — uniform downward acceleration. Two independent stories married by a shared clock. That marriage is the entire reason the parametric viewpoint is the right language here: it keeps the across-motion and the up-motion cleanly separate while describing one curving arc.

When does the shell land, and how far?

The shell returns to the ground when $y(t) = 0$ a second time. Factor the vertical equation:

$$(v_0\sin\theta)\,t - \tfrac{1}{2}g t^2 = t\!\left(v_0\sin\theta - \tfrac{1}{2}g t\right) = 0.$$

The two roots are $t = 0$ (the muzzle) and the landing time

$$t_{\text{land}} = \frac{2 v_0 \sin\theta}{g}.$$

Feed that time into the horizontal equation to get the range — the distance downrange where the shell strikes:

$$R(\theta) = (v_0\cos\theta)\cdot\frac{2 v_0\sin\theta}{g} = \frac{v_0^2\,(2\sin\theta\cos\theta)}{g} = \frac{v_0^2 \sin(2\theta)}{g},$$

using the double-angle identity $2\sin\theta\cos\theta = \sin(2\theta)$. This single tidy formula is the officer's whole world. With $v_0 = 250$ and $g = 9.8$, the coefficient $v_0^2/g = 62500/9.8 \approx 6378$ meters. So

$$R(\theta) \approx 6378\,\sin(2\theta)\ \text{meters}.$$

Is the target reachable? The largest $R$ can ever be is $6378$ m, achieved when $\sin(2\theta) = 1$. The target sits at $6000$ m — inside that ceiling, so yes, it can be hit. (Had the target been at $7000$ m, no elevation on Earth would reach it at this muzzle speed, and the only answer is a bigger charge.)

Finding the elevation

Set $R(\theta) = 6000$ and solve for $\theta$:

$$6378\,\sin(2\theta) = 6000 \quad\Longrightarrow\quad \sin(2\theta) = 0.9407.$$

Now $2\theta = \arcsin(0.9407)$. The arcsine has two solutions in $[0, 180^\circ]$, and that fact is the heart of gunnery:

$$2\theta = 70.2^\circ \quad\text{or}\quad 2\theta = 180^\circ - 70.2^\circ = 109.8^\circ,$$

so

$$\theta \approx 35.1^\circ \quad\text{or}\quad \theta \approx 54.9^\circ.$$

Two elevations hit the same target. The low one, $35.1^\circ$, sends a flat, fast shell that arrives quickly — useful against a moving target. The high one, $54.9^\circ$, lobs the shell up and over, arriving steeply from above — useful for dropping fire behind a ridge the flat shot would smash into. They are a complementary pair: $35.1^\circ + 54.9^\circ = 90^\circ$, exactly as $\sin(2\theta) = \sin(180^\circ - 2\theta)$ predicts. Every artillery manual distinguishes "low-angle" from "high-angle" fire for precisely this reason, and the distinction is a one-line consequence of the parametric model.

Why $45^\circ$ is special

Suppose instead the officer simply wants the farthest possible shot — the maximum range, target unknown. This is an optimization problem (Chapter 10) sitting on top of the parametric model. We maximize $R(\theta) = \dfrac{v_0^2}{g}\sin(2\theta)$ over $\theta \in [0^\circ, 90^\circ]$. Since $v_0^2/g$ is a constant, we just maximize $\sin(2\theta)$, which peaks at $\sin(2\theta) = 1$, i.e. $2\theta = 90^\circ$, so

$$\boxed{\theta = 45^\circ}, \qquad R_{\max} = \frac{v_0^2}{g} \approx 6378\ \text{m}.$$

The optimal elevation for maximum range in a vacuum is exactly $45^\circ$. It is worth pausing on how cleanly this falls out: a fact every shot-putter and golfer half-knows by feel becomes a single derivative-free maximization once the trajectory is written parametrically.

The shape of the path

The officer also wants to know whether the shell clears a 40-meter ridge that stands 3,000 meters out. For that he needs the shape of the arc, not just where it lands — and shape means eliminating the parameter. Solve the horizontal equation for time, $t = x/(v_0\cos\theta)$, and substitute into the vertical:

$$y = (v_0\sin\theta)\cdot\frac{x}{v_0\cos\theta} - \frac{g}{2}\left(\frac{x}{v_0\cos\theta}\right)^2 = x\tan\theta - \frac{g}{2v_0^2\cos^2\theta}\,x^2.$$

This is $y = (\text{const})x - (\text{const})x^2$ — a downward parabola, the conic section we will study in full in Chapter 27. Newton's celebrated conclusion, recovered in three lines: a projectile under uniform gravity flies a parabolic arc. For the high-angle solution $\theta = 54.9^\circ$ ($\tan\theta \approx 1.42$, $\cos^2\theta \approx 0.329$), the coefficient of $x^2$ is $\dfrac{9.8}{2(62500)(0.329)} \approx 2.38\times 10^{-4}$, so

$$y \approx 1.42\,x - 2.38\times 10^{-4}\,x^2.$$

At the ridge, $x = 3000$: $y \approx 4260 - 2142 = 2118$ m — the high arc clears the 40-meter ridge with thousands of meters to spare. The low-angle shot ($\theta = 35.1^\circ$, $\tan\theta \approx 0.703$) gives at $x = 3000$ a height of $y \approx 2109 - 1071 \approx 1038$ m, also clearing it. Both shots are geometrically valid; the choice is tactical.

Where the slope is zero: the apex

One more reading of the model. The shell is at its highest exactly where the tangent to the path is horizontal — where the §25.3 condition $\dot y = 0$ holds. Since $\dot y = v_0\sin\theta - g t$, the apex occurs at $t = v_0\sin\theta / g$, precisely half the flight time, as symmetry demands. The maximum height is

$$y_{\max} = \frac{v_0^2 \sin^2\theta}{2g}.$$

For the high-angle shot this is $\dfrac{62500 \cdot 0.671}{19.6} \approx 2140$ m — consistent with the parabola we just plotted. The slope formula $dy/dx = \dot y/\dot x$ and the landing-time algebra are telling the same story from two directions, which is exactly the cross-check a careful modeler wants.

The honest caveat

Everything above lives in a vacuum. A real shell pushes through air, and the drag force — roughly proportional to the square of speed — bleeds energy continuously. The true trajectory is no longer a perfect parabola: it is steeper on the way down than on the way up, the range falls well short of $v_0^2/g$, and the optimal angle for maximum range drops below $45^\circ$ (typically to the high-30s for artillery, and far lower for a long-jumper who simply cannot generate launch speed at $45^\circ$). Real firing tables encode these corrections empirically, shot by shot. But the parametric skeleton never changes: two coupled rate laws sharing a clock. The vacuum model is not wrong so much as it is the clean first term of a more elaborate story — and it is the term that tells you, in one line, whether a target is even reachable.

Discussion Questions

1. The two-angle phenomenon. Explain in physical terms why a low-angle and a high-angle shot reach the same range. Which arrives faster, and which arrives more steeply? Tie your answer to the identity $\sin(2\theta) = \sin(180^\circ - 2\theta)$.

2. Reachability. Derive the condition on $v_0$ for a target at distance $D$ to be reachable at all. What is the unique firing angle in the borderline case where the target sits exactly at maximum range?

3. Apex via two methods. Find the time and height of the apex (a) by setting $\dot y = 0$ and (b) by completing the square in the parabolic Cartesian equation $y = x\tan\theta - cx^2$. Confirm they agree.

4. Launching from a height. Suppose the gun sits on a cliff of height $h$, so $y(t) = h + (v_0\sin\theta)t - \tfrac12 g t^2$. Argue qualitatively why the maximum-range angle is now less than $45^\circ$, and sketch how you would set up the optimization (cf. Exercise G4).

5. The drag correction. Air resistance lowers both the range and the optimal angle. Without solving the harder equations, explain why the downward leg of a real trajectory is steeper than the upward leg.

A Short Annotated Reading

• Halliday, Resnick & Walker, Fundamentals of Physics (Wiley). The standard introductory treatment of projectile motion; its chapter on motion in two dimensions develops exactly the $x(t)$, $y(t)$ decomposition used here, with the range and apex formulas worked in full.

• Marion & Thornton, Classical Dynamics of Particles and Systems (Cengage). Goes one level deeper, adding linear and quadratic air resistance and showing how the optimal angle slides below $45^\circ$ — the rigorous version of this case study's closing caveat.

• Hahn, A. J., Basic Calculus: From Archimedes to Newton to Its Role in Science (Springer, 1998). Traces the historical arc from Galileo's discovery that the path is a parabola to Newton's calculus, situating the §25.6 model in the story of how calculus and ballistics grew up together.