Projectile motion with resistance in HSC Mathematics Extension 2: resolving motion into components and modelling air resistance proportional to velocity in two dimensions

What this dot point is asking

NESA wants you to analyse the motion of a projectile using calculus. You resolve the motion into independent horizontal and vertical components, set up the equations from Newton's second law, and integrate. You must handle both the resistance-free case (constant horizontal velocity, constant vertical acceleration) and the case of a resistance proportional to velocity acting in both directions.

Resolving into components

Projectile motion is two-dimensional, but the horizontal and vertical directions can be treated as separate one-dimensional problems linked only by a shared time $t$. With initial speed $V$ at angle $\theta$ to the horizontal, the initial velocity components are

You set up an equation of motion in each direction, integrate to get velocity, integrate again to get position, and apply initial conditions at each stage.

How resistance reshapes the path, stage by stage

The clearest way to see what resistance does is to draw the resisted trajectory against the resistance-free parabola. The panels below set up the launch, draw the ideal parabola, then overlay the resisted path and read off the differences.

Stage 1, resolve the launch velocity. Whether or not there is resistance, the first step is the same: split the launch speed $V$ at angle $\theta$ into a horizontal part $V\cos\theta$ and a vertical part $V\sin\theta$. These are the initial conditions for the two component equations.

Stage 2, the resistance-free parabola. With gravity the only force, the horizontal velocity is constant and the vertical motion is uniformly accelerated, so the path is the familiar symmetric parabola. This is the benchmark the resisted motion is compared against.

Stage 3, overlay the resisted path. Resistance opposes the velocity throughout, bleeding off speed in both directions. The resisted trajectory (solid) sits entirely below the parabola (faint): it reaches a lower peak and lands well short, because the horizontal velocity is decaying instead of staying constant.

Stage 4, read off the asymmetry. Unlike the parabola, the resisted path is not symmetric: the descent is steeper than the ascent, and the highest point is past the midpoint of the (shorter) range. The horizontal velocity decays towards zero, so the horizontal distance approaches a finite bound rather than growing without limit. These three features, lower, shorter and asymmetric, are exactly what a sketch question rewards.

Projectile motion without resistance

With gravity as the only force, taking up as positive,

Integrating the horizontal equation gives constant velocity $\dot{x} = V\cos\theta$ and position $x = (V\cos\theta)\,t$. Integrating the vertical equation twice gives $\dot{y} = V\sin\theta - gt$ and

These are the familiar equations: the path is a parabola, and the range, greatest height and time of flight follow by setting the appropriate quantity to zero.

Projectile motion with resistance

When a resistance of magnitude proportional to speed acts opposite to the velocity, it splits into components along each axis. The equations become

where $k > 0$. Each is a first-order linear equation in the relevant velocity component. The horizontal equation $m\dfrac{d\dot{x}}{dt} = -k\dot{x}$ separates to give an exponential decay,

so the horizontal velocity decays toward zero and the horizontal range is bounded. The vertical equation has a terminal-velocity structure like resisted rectilinear motion: the upward motion decelerates faster than under gravity alone, and the descent approaches a terminal speed.

Strategy

Always resolve first, then integrate each component separately, applying initial conditions immediately after each integration so constants are pinned down early. The shared parameter is time $t$; eliminate it at the end if a relation between $x$ and $y$ (the trajectory) is required.

How exam questions ask about projectile motion with resistance

The wording signals which component and which integration are wanted:

• "Show that the initial velocity is $\mathbf{v}(0) = (V\cos\theta, V\sin\theta)$." A one-mark resolution; quote both components.
• "The resistance is proportional to the velocity. Show that $\mathbf{v}(t) = \ldots$" Write $m\ddot{x} = -k\dot{x}$ and $m\ddot{y} = -mg - k\dot{y}$, solve each first-order equation in the velocity, and apply the initial components. The horizontal one separates to an exponential; the vertical one is linear with a constant (terminal) part plus a transient.
• "Find the horizontal range / show the range is bounded." Integrate the decaying $\dot{x}$ to get $x(t)$; as $t \to \infty$ the horizontal distance approaches a finite limit, which is the bound.
• "Find the greatest height / time to the top." Set the vertical velocity $\dot{y} = 0$ and solve for $t$, exactly as without resistance but with the resisted $\dot{y}$.
• "Find the terminal velocity of the descent." Set $\ddot{y} = 0$ in the vertical equation, giving the steady descent speed.
• "Find the speed / direction at time $t$." Combine the components: speed $\sqrt{\dot{x}^2 + \dot{y}^2}$, direction $\arctan(\dot{y}/\dot{x})$, watching the sign of $\dot{y}$ after the top.
• "For the resistance-free case ... find range / height / time of flight." The standard parametric formulas apply; resistance questions usually build on a resistance-free part first.
• "When is the position vector perpendicular to the velocity vector?" Form $\mathbf{r}\cdot\mathbf{v} = 0$ and solve, as in the 2021 question; this is a vectors-and-calculus crossover.