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.

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 4.76 min answerUpdated 2026-05-29

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

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.

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.

Worked example

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

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

Vertical motion: $y = 10t - \tfrac{1}{2}(10)t^2 = 10t - 5t^2$ . The projectile lands when $y = 0$ :

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

Horizontal motion: $x = 10\sqrt{3}\,t$ , so the range is

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

Check using the standard range formula $\dfrac{V^2\sin 2\theta}{g} = \dfrac{400\sin 60^\circ}{10} = 40 \times \tfrac{\sqrt{3}}{2} = 20\sqrt{3}$ . Agreement confirms the result.

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.