Projectile motion with air drag: how quadratic drag changes the flight

The quadratic drag model

For a projectile moving quickly through air, the drag force is usually modeled as proportional to the square of the speed: F_d = k·v², pointing opposite to the velocity vector. The constant k lumps together the air density ρ, the drag coefficient C_d and the cross-sectional area A, so k = ½·ρ·C_d·A.

With that force, Newton's second law gives an acceleration of −g·ŷ − (k/m)·|v|·v. The ratio k/m is the key parameter: it has units of inverse length, and it decides how quickly the drag term grows relative to gravity. Doubling the mass halves k/m, which is why a heavy ball is deflected less than a light one of the same size.

The equations no longer have a simple closed-form solution, so the calculator integrates them numerically with fourth-order Runge–Kutta and refines the landing point where the height crosses zero. The vacuum trajectory is computed analytically at the same time, which gives a built-in comparison.

Why drag changes the shape of the flight

Drag removes mechanical energy continuously, so the projectile cannot rise as high or travel as far as it would in vacuum. The horizontal speed falls throughout the flight instead of staying constant, and the vertical speed also decays toward the fall, which limits the speed at impact.

Because drag increases with speed, it acts most strongly where the projectile moves fastest. The apex is shifted earlier than the midpoint, the descent is steeper than the climb, and the impact angle is larger than the launch angle. The symmetric parabola of the ideal model becomes an asymmetric arc that is flatter on the way up.

One consequence surprises many people: the drag trajectory can have a shorter flight time than the vacuum case, even though it also rises less. Gravity needs less time to bring down an object whose vertical speed has already been reduced by drag.

Worked example: 20 m/s at 45 degrees

Take a 0.5 kg projectile launched at 20 m/s and 45 degrees from ground level, with gravity 9.81 m/s² and a drag coefficient k = 0.005 kg/m. Then k/m = 0.01 m⁻¹. The ideal vacuum solution gives a range of about 40.77 m, a flight time of 2.88 s and a maximum height of 10.19 m.

With drag, the same launch gives a range of about 31.32 m, a flight time of 2.67 s, a maximum height of 8.78 m and an impact speed of 15.72 m/s. The range is roughly 23% shorter than the vacuum value, and the impact speed is far below the vacuum impact speed of 20 m/s.

Setting k = 0 in the calculator reproduces the vacuum numbers exactly, which is a useful check: the numerical integrator and the analytic formula should agree when the drag term is switched off.

Terminal velocity as a feel for the effect

When gravity and drag balance, a falling object stops accelerating at its terminal velocity. For quadratic drag the balance gives v_t = √(m·g ÷ k). In the worked example, v_t = √(0.5 × 9.81 ÷ 0.005) ≈ 31.3 m/s.

That number is a useful scale: projectiles moving well below their terminal velocity behave almost ideally, while those approaching or exceeding it are strongly affected by drag. The ratio of launch speed to terminal velocity, together with k/m and the flight distance, tells you quickly whether an ideal calculation is defensible.

The same relation explains why small, light objects such as leaves, feathers and dust never follow textbook parabolas: their terminal velocity is very low, so the air dominates their motion almost immediately.

When the ideal model is still good enough

For a dense, compact projectile over a short flight at modest speed, drag may change the range by only a small percentage, and the vacuum model is a reasonable first estimate. For long flights, light objects, high launch speeds or precise predictions, the drag calculation is the better starting point.

The main uncertainty is usually k itself. Air density varies with altitude and weather, the drag coefficient depends on shape and orientation, and real projectiles may tumble or spin. A drag model is therefore a structured way to understand the correction, not a substitute for measurement when the answer matters.