What is sign of gravity and resistance?

With up positive, gravity contributes -mg and resistance opposes velocity in each direction. A misplaced sign reverses the physics.

§-Syllabus dot point

NSWMaths Extension 2Syllabus dot point

How is two-dimensional projectile motion analysed with calculus, and how does air resistance proportional to velocity change the horizontal and vertical motion?

Analyse projectile motion using calculus, resolving into horizontal and vertical components, and extend to projectiles experiencing a resistance proportional to velocity

A focused answer to the HSC Maths Extension 2 dot point on projectile motion. Resolving into independent horizontal and vertical equations, projectile motion without resistance, and resistance proportional to velocity in two dimensions, with verified worked examples.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Resolving into components
Projectile motion without resistance
Projectile motion with resistance
Strategy
How exam questions ask about projectile motion with resistance

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

\dot{x}(0) = V\cos\theta, \qquad \dot{y}(0) = V\sin\theta.

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,

\ddot{x} = 0, \qquad \ddot{y} = -g.

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

y = (V\sin\theta)\,t - \tfrac{1}{2}gt^2.

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

m\ddot{x} = -k\dot{x}, \qquad m\ddot{y} = -mg - k\dot{y},

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,

\dot{x} = V\cos\theta\, e^{-kt/m},

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.

A projectile is fired at

V = 20

m/s at

\theta = 30^\circ

with no resistance. Taking

g = 10

, find the time of flight and the horizontal range

Find the time of flight and horizontal range for a projectile with no resistance

The two components of motion are independent. Horizontal velocity is constant (no resistance) and vertical motion follows constant acceleration under gravity. We resolve, then integrate each component separately.

Resolve the initial velocity into components: Split the launch speed into horizontal and vertical parts using the launch angle:

\dot{x}(0) = 20\cos 30^\circ = 20 \times \tfrac{\sqrt{3}}{2} = 10\sqrt{3}, \qquad \dot{y}(0) = 20\sin 30^\circ = 20 \times \tfrac{1}{2} = 10.

Find the time of flight from the vertical equation: With $\ddot{y} = -g = -10$ , integrating twice and applying initial conditions gives $y = 10t - 5t^2$ . The projectile returns to the ground when $y = 0$ :

10t - 5t^2 = 0 \quad\Longrightarrow\quad 5t(2 - t) = 0 \quad\Longrightarrow\quad t = 2 \text{ s.}

The non-zero root gives the time of flight. The root $t = 0$ corresponds to the launch instant.

Find the horizontal range: Since horizontal acceleration is zero, horizontal velocity remains constant at $\dot{x}(0) = 10\sqrt{3}$ . The horizontal range at $t = 2$ is:

x = 10\sqrt{3} \times 2 = 20\sqrt{3} \approx 34.6 \text{ m.}

Verify with the standard range formula: The range formula $\dfrac{V^2\sin 2\theta}{g}$ gives $\dfrac{400\sin 60^\circ}{10} = 40 \times \tfrac{\sqrt{3}}{2} = 20\sqrt{3}$ , which agrees with the calculated value.

Final answer: time of flight $= 2$ s; horizontal range $= 20\sqrt{3} \approx 34.6$ m.

With resistance: the horizontal velocity and the bound on the range

A unit-mass particle is launched from the origin with horizontal component $\dot{x}(0) = 20\sqrt{3}$ and meets a resistance giving horizontal equation $\ddot{x} = -4\dot{x}$ (the 2023 setup). We find the horizontal velocity, the horizontal distance, and the limit the range cannot exceed.

Solve the horizontal velocity equation: The horizontal equation is first order in $\dot{x}$ . Writing $\dot{x} = w$ , $\dfrac{dw}{dt} = -4w$ , so $\dfrac{1}{w}\,dw = -4\,dt$ and $\ln w = -4t + C$ , giving $w = A e^{-4t}$ . At $t = 0$ , $w = 20\sqrt{3}$ , so $A = 20\sqrt{3}$ and

\dot{x} = 20\sqrt{3}\,e^{-4t}.

The horizontal velocity decays exponentially, never reaching zero but getting arbitrarily small.

Integrate to the horizontal position: Integrate $\dot{x} = 20\sqrt{3}\,e^{-4t}$ with respect to $t$ :

x = -\frac{20\sqrt{3}}{4}e^{-4t} + D = -5\sqrt{3}\,e^{-4t} + D.

At $t = 0$ , $x = 0$ , so $0 = -5\sqrt{3} + D$ , giving $D = 5\sqrt{3}$ . Therefore $x = 5\sqrt{3}\big(1 - e^{-4t}\big)$ .

Find the bound on the horizontal range: As $t \to \infty$ , $e^{-4t} \to 0$ , so $x \to 5\sqrt{3}$ . The horizontal distance can never exceed $5\sqrt{3}$ metres, however long the flight lasts. This is the bounded-range feature that resistance forces and that the resistance-free parabola does not have.
Compare with the resistance-free range: Without resistance the same horizontal launch would travel $20\sqrt{3} \approx 34.6$ m in the $2$ s flight. With resistance the horizontal reach is capped at $5\sqrt{3} \approx 8.7$ m, dramatically shorter, which is the quantitative form of the "lands well short" feature in the stage-by-stage figures.
Final answer: $\dot{x} = 20\sqrt{3}\,e^{-4t}$ and $x = 5\sqrt{3}\big(1 - e^{-4t}\big)$ , so the horizontal range is bounded above by $5\sqrt{3}$ m.

Common traps

Mixing the components: Horizontal and vertical motions are independent; gravity and the vertical resistance term never appear in the horizontal equation. Keep the two equations separate.
Sign of gravity and resistance: With up positive, gravity contributes $-mg$ and resistance opposes velocity in each direction. A misplaced sign reverses the physics.
Applying constant-acceleration formulas when there is resistance: The formulas $y = ut - \tfrac{1}{2}gt^2$ hold only without resistance. With resistance you must integrate the differential equations.
Forgetting the angle resolution: The initial speed must be split as $V\cos\theta$ and $V\sin\theta$ . Using $V$ directly in one direction is wrong unless the launch is purely horizontal or vertical.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC3 marksA particle of mass 1 kg is projected from the origin with speed 40 m s^-1 at an angle 30 degrees to the horizontal plane. (i) Show that the initial velocity is v(0) = (20 sqrt(3), 20). The forces are gravity and air resistance, where the air resistance is proportional to the velocity vector with constant of proportionality 4, and g = 10 m s^-2. (ii) Show that v(t) = (20 sqrt(3) e^(-4t), (45/2) e^(-4t) - 5/2).

Show worked answer →

Part (i). Resolve the launch velocity into components. Horizontal: 40 cos 30 = 40 (sqrt(3)/2) = 20 sqrt(3). Vertical: 40 sin 30 = 40 (1/2) = 20. So v(0) = (20 sqrt(3), 20).

Part (ii). With mass 1, Newton's second law gives a = -4 v + g, where g = (0, -10). Resolving into components (the two directions are independent):

Horizontal: x'' = -4 x'. Writing x' = dx'/dt, dx'/x' = -4 dt, so ln x' = -4t + C, x' = A e^(-4t). At t = 0, x' = 20 sqrt(3), so A = 20 sqrt(3): x' = 20 sqrt(3) e^(-4t).

Vertical: y'' = -4 y' - 10. This is linear; the steady part is y' = -10/4 = -5/2, and the transient is B e^(-4t), so y' = -5/2 + B e^(-4t). At t = 0, y' = 20, so 20 = -5/2 + B, giving B = 45/2: y' = (45/2) e^(-4t) - 5/2.

Hence v(t) = (20 sqrt(3) e^(-4t), (45/2) e^(-4t) - 5/2).

Mark notes: 1 mark for resolving to give v(0); then 1 mark for solving the horizontal equation and 1 mark for solving the vertical equation with the correct constant.

2021 HSC5 marksA particle projected from the origin with initial speed u m s^-1 at angle theta to the positive x-axis lands on the x-axis. Its position vector is r(t) = (u t cos theta, -g t^2/2 + u t sin theta). For some value(s) of theta there are two times during the flight when the position vector is perpendicular to the velocity vector. Find the value(s) of theta for which this occurs, justifying that both times occur during the time of flight.

Show worked answer →

The velocity is v(t) = r'(t) = (u cos theta, u sin theta - g t). Perpendicularity means r . v = 0.

r . v = u t cos theta (u cos theta) + (u t sin theta - g t^2/2)(u sin theta - g t).

Expanding and using cos^2 + sin^2 = 1, the bracketed terms combine to

r . v = u^2 t - (3/2) u g t^2 sin theta + (1/2) g^2 t^3.

Take out t (with t > 0 during flight) and multiply by 2:

g^2 t^2 - 3 u g sin theta . t + 2 u^2 = 0.

For TWO times we need two distinct positive roots. Discriminant > 0: 9 u^2 g^2 sin^2 theta - 8 g^2 u^2 > 0, so sin^2 theta > 8/9, that is sin theta > 2 sqrt(2)/3 (theta acute).

Both roots are positive: sum = 3 u sin theta/g > 0 and product = 2 u^2/g^2 > 0.

Both roots lie before landing (time of flight T = 2 u sin theta/g): the larger root is (u/(2g))(3 sin theta + sqrt(9 sin^2 theta - 8)), and this is less than T = (u/(2g))(4 sin theta) provided sqrt(9 sin^2 theta - 8) < sin theta, i.e. 9 sin^2 theta - 8 < sin^2 theta, i.e. sin^2 theta < 1, which holds for theta < pi/2.

Conclusion: arcsin(2 sqrt(2)/3) < theta < pi/2 (about 70.5 degrees to 90 degrees).

Mark notes: 1 mark for v(t), 1 mark for forming r . v = 0, 1 mark for the quadratic in t, 1 mark for the discriminant condition sin theta > 2 sqrt(2)/3, 1 mark for justifying both roots fall within the time of flight.