What is trajectory (Cartesian) equation?

Eliminate t from the position equations. From x = V cos α · t, t = (x)/(V cos α). Substitute into y:

What is range on inclined ground?

For ground inclined at angle β to the horizontal, the projectile lands when its path meets the line of the slope, so you set y(t) = x(t)tanβ (sign depending on uphill or downhill) and solve for t. This is a harder Section II application; the method is unchanged, only the landing condition is.

What is maximum height?

set y(t) = 0, giving t = (V sin α)/(g), then substitute.

What is wrong angle in sin 2 α?

The range formula uses sin 2 α, not sin^2 α. The double angle comes from 2sin α cos α = sin 2 α.

§-Syllabus dot point

NSWMaths Extension 1Syllabus dot point

How do we model projectile motion in two dimensions, and what quantities (range, maximum height, time of flight) can we extract?

Model projectile motion in two dimensions using parametric equations and find range, maximum height, time of flight and trajectory equation

A focused answer to the HSC Maths Extension 1 dot point on projectile motion. The parametric equations for position, velocity and acceleration under gravity, the Cartesian trajectory equation, and standard quantities (range, maximum height, time of flight), with worked examples.

Generated by Claude Opus 4.813 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

NESA wants you to model a projectile launched from ground level (or some height) with a given initial speed and angle, write down the parametric position equations, find the trajectory equation, and compute the standard quantities: range, time of flight, maximum height, and velocity at a given time.

The whole topic rests on a single physical idea: horizontal and vertical motion are independent. Gravity acts only downward, so it changes the vertical velocity and leaves the horizontal velocity untouched. That is why a problem in two dimensions splits cleanly into two one-dimensional problems that share the same clock $t$ .

The answer

The model

A projectile launched from the origin with initial speed $V$ at angle $\alpha$ above the horizontal, with gravitational acceleration $g$ downward and no air resistance:

Velocity components:

\dot x(t) = V \cos \alpha, \qquad \dot y(t) = V \sin \alpha - g t.

Position equations:

x(t) = V \cos \alpha \cdot t, \qquad y(t) = V \sin \alpha \cdot t - \tfrac{1}{2} g t^2.

Acceleration:

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

Horizontal motion is at constant velocity; vertical motion is under constant downward acceleration $g$ .

Building the trajectory, stage by stage

The figures below build a launch from the ground one feature at a time, in the order you should set the problem up.

Stage 1, resolve the launch velocity. The initial velocity $V$ at angle $\alpha$ splits into a horizontal part $V\cos\alpha$ and a vertical part $V\sin\alpha$ . Every later formula is built from these two components, so resolve first.

Stage 2, trace the path. Constant horizontal velocity carries the projectile sideways at a steady rate while gravity bends the vertical motion, and the combination is a parabola. Eliminating $t$ from the two position equations gives its Cartesian form, $y = x\tan\alpha - \dfrac{gx^2}{2V^2\cos^2\alpha}$ .

Stage 3, find the maximum height. At the top of the path the projectile is momentarily moving horizontally, so the vertical velocity is zero: $\dot y = 0$ . Solving $V\sin\alpha - gt = 0$ gives $t = \dfrac{V\sin\alpha}{g}$ ; substituting into $y$ gives $H = \dfrac{V^2\sin^2\alpha}{2g}$ .

Stage 4, find the range and time of flight. On level ground the projectile lands when $y = 0$ again, at $t = \dfrac{2V\sin\alpha}{g}$ - exactly twice the time to the top, by symmetry. The range is the horizontal distance covered in that time, $R = V\cos\alpha\cdot t = \dfrac{V^2\sin 2\alpha}{g}$ .

Standard derived quantities

Time of flight (on horizontal ground): set $y(t) = 0$ .

T = \frac{2 V \sin \alpha}{g}.

Range on horizontal ground: $R = V \cos \alpha \cdot T$ .

R = \frac{V^2 \sin 2 \alpha}{g}.

(Maximum range occurs at $\alpha = 45^\circ$ , because $\sin 2\alpha$ peaks there.)

Maximum height: set $\dot y(t) = 0$ , giving $t = \frac{V \sin \alpha}{g}$ , then substitute.

H = \frac{V^2 \sin^2 \alpha}{2 g}.

Trajectory (Cartesian) equation

Eliminate $t$ from the position equations. From $x = V \cos \alpha \cdot t$ , $t = \frac{x}{V \cos \alpha}$ . Substitute into $y$ :

y = x \tan \alpha - \frac{g x^2}{2 V^2 \cos^2 \alpha} = x \tan \alpha - \frac{g x^2 (1 + \tan^2 \alpha)}{2 V^2}.

This is a downward parabola. The second form, using $\sec^2\alpha = 1 + \tan^2\alpha$ , is handy when a question gives or asks for the launch angle through $\tan\alpha$ .

Launched from a height

If the projectile starts at height $h$ , then $y(t) = h + V \sin \alpha \cdot t - \tfrac{1}{2} g t^2$ . The time of flight is now the larger root of this quadratic in $t$ , and the symmetric range formula no longer applies - you must solve the quadratic. This is the single most common way the standard formulas are made to fail, so check the launch height before reaching for $R = \frac{V^2\sin 2\alpha}{g}$ .

Velocity: speed and direction

Magnitude of velocity at time $t$ :

v(t) = \sqrt{(\dot x)^2 + (\dot y)^2}.

Direction (above horizontal):

\theta(t) = \arctan \frac{\dot y}{\dot x}.

The speed is minimum at the top of the trajectory, where $\dot y = 0$ and the speed is just $V\cos\alpha$ . On level ground the landing speed equals the launch speed, by symmetry of the parabola.

Range on inclined ground

For ground inclined at angle $\beta$ to the horizontal, the projectile lands when its path meets the line of the slope, so you set $y(t) = \pm x(t)\tan\beta$ (sign depending on uphill or downhill) and solve for $t$ . This is a harder Section II application; the method is unchanged, only the landing condition is.

How exam questions ask about projectile motion

"Show that the position is given by ..." / "Derive the equations of motion." Start from $\ddot x = 0$ , $\ddot y = -g$ and integrate twice, using the initial velocity components and launch point as the constants. Show the integration; do not just quote the result.
"Find the greatest height." A $\dot y = 0$ question: solve for $t$ , substitute into $y$ .
"Find the range" or "how far from the launch point does it land?" A $y = 0$ question (level ground) or a "meets the slope / cliff base" condition; solve for $t$ , then substitute into $x$ .
"Find the speed and direction after $t$ seconds / on landing." Combine the components: speed $\sqrt{\dot x^2 + \dot y^2}$ , angle $\arctan(\dot y/\dot x)$ . Mind the sign of $\dot y$ on the way down.
"Show the trajectory is a parabola" or "find the Cartesian equation." Eliminate $t$ between the two position equations.
"For what angle is the range greatest?" Maximise $\sin 2\alpha$ : the answer is $45^\circ$ on level ground.
"The projectile is launched from a height $h$ ..." Add $h$ to $y(t)$ and solve the resulting quadratic; the symmetric formulas do not apply.

Worked example

Apply the range formula

A ball is kicked from ground level at $V = 20$ m/s at $\alpha = 30^\circ$ , with $g = 9.8$ m/s $^2$ . The standard range formula is $R = \frac{V^2 \sin 2\alpha}{g}$ . Here $V^2 = 400$ and $\sin 60^\circ \approx 0.866$ .

Substitute and evaluate:

R = \frac{400 \cdot \sin 60^\circ}{9.8} = \frac{400 \cdot 0.866}{9.8} \approx 35.3 \text{ m}.

Final answer: approximately $35.3$ m.

Apply the maximum-height formula

With the same ball ( $V = 20$ m/s, $\alpha = 30^\circ$ ), use $H = \frac{V^2 \sin^2 \alpha}{2g}$ . Note $\sin 30^\circ = 0.5$ , so $\sin^2 30^\circ = 0.25$ .

H = \frac{400 \cdot 0.25}{2 \cdot 9.8} = \frac{100}{19.6} \approx 5.10 \text{ m}.

Final answer: approximately $5.10$ m.

Evaluate the vertical position at a given time

The vertical position is $y(t) = V \sin\alpha \cdot t - \frac{1}{2} g t^2$ . Substituting $V = 20$ , $\alpha = 30^\circ$ , $g = 9.8$ , and $t = 1.5$ :

y(1.5) = 20 \sin 30^\circ \cdot 1.5 - \tfrac{1}{2} \cdot 9.8 \cdot 1.5^2 = 15 - 11.025 = 3.975 \text{ m}.

Final answer: approximately $4.0$ m. (Since the time to the top is $\frac{V\sin\alpha}{g} = \frac{10}{9.8}\approx 1.02$ s, at $t = 1.5$ s the ball is already past its peak and on the way down.)

Find the trajectory equation

The Cartesian trajectory is $y = x \tan\alpha - \frac{g x^2}{2 V^2 \cos^2\alpha}$ . For $V = 25$ m/s and $\alpha = 40^\circ$ , so $V^2 = 625$ , $\tan 40^\circ \approx 0.839$ , and $\cos^2 40^\circ \approx 0.587$ .

y \approx 0.839 x - \frac{9.8 \, x^2}{2 \cdot 625 \cdot 0.587} \approx 0.839 x - 0.0134 \, x^2.

Final answer: $y \approx 0.839x - 0.0134 x^2$ , a downward parabola.

Find the landing speed by symmetry

For a projectile launched and landing on level ground with $V = 15$ m/s and $\alpha = 60^\circ$ , the trajectory is symmetric. The vertical speed component is equal in magnitude at launch and landing (gravity removes then restores it symmetrically), and the horizontal component is unchanged throughout. Therefore the speed at landing equals the launch speed.

Final answer: the landing speed is $15$ m/s, directed at $60^\circ$ below the horizontal.

Find the time of flight from a cliff

A stone is thrown from a $25$ m cliff at $V = 20$ m/s, $\alpha = 30^\circ$ , $g = 9.8$ . Because it starts above the landing level, the symmetric formula does not apply. The vertical position measured from the launch point is $y(t) = 20\sin 30^\circ\, t - \tfrac12(9.8)t^2 = 10t - 4.9t^2$ , and the stone lands when $y = -25$ (the base of the cliff):

10t - 4.9t^2 = -25 \implies 4.9t^2 - 10t - 25 = 0.

Solve the quadratic and take the positive root:

t = \frac{10 + \sqrt{100 + 4\cdot 4.9\cdot 25}}{2\cdot 4.9} = \frac{10 + \sqrt{590}}{9.8} \approx \frac{10 + 24.29}{9.8} \approx 3.50 \text{ s}.

Final answer: the stone is in the air for about $3.50$ s, longer than the $\frac{2V\sin\alpha}{g}\approx 2.04$ s the symmetric formula would have (wrongly) given - the cliff height is exactly what the symmetric formula ignores.

Find the angle for maximum range

The range on level ground is $R = \frac{V^2 \sin 2\alpha}{g}$ . For fixed $V$ and $g$ , this is maximised when $\sin 2\alpha$ is as large as possible. The maximum of $\sin$ is $1$ , achieved at $2\alpha = 90^\circ$ .

2\alpha = 90^\circ \implies \alpha = 45^\circ.

Final answer: the maximum range is achieved at a launch angle of $45^\circ$ .

Common traps

Sign of gravity: With upward positive, gravitational acceleration is $-g$ . Mixing the sign in the position equation flips the parabola.
Confusing initial speed with a component: $V \cos \alpha$ is the horizontal component, $V \sin \alpha$ the vertical. Swapping them swaps the trajectory.
Wrong angle in $\sin 2 \alpha$: The range formula uses $\sin 2 \alpha$ , not $\sin^2 \alpha$ . The double angle comes from $2\sin \alpha \cos \alpha = \sin 2 \alpha$ .
Forgetting a launch height: Many HSC questions launch from a cliff or table. Add the offset $h$ to $y(t)$ and solve the quadratic; the symmetric range and time formulas no longer hold.
Using a max-height condition for range, or vice versa: Maximum height is $\dot y = 0$ ; landing is $y = 0$ . Pick the right condition.
Mixing units: Use SI ( $g = 9.8$ m/s $^2$ , or $g = 10$ m/s $^2$ only if the question allows it). State $g$ and check the question's value.

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.

2022 HSC Q214 marksA particle is projected from ground level at angle

\alpha

above the horizontal with initial speed

V

. Assuming acceleration due to gravity is

g

, derive expressions for the range

R

on horizontal ground and the maximum height

H

Show worked answer →

Components of initial velocity: $V_x = V \cos \alpha$ , $V_y = V \sin \alpha$ .

Position equations (with the origin at launch and upward positive):

$x(t) = V \cos \alpha \cdot t$

$y(t) = V \sin \alpha \cdot t - \frac{1}{2} g t^2$

Range: set $y(t) = 0$ to find time of flight. $t = 0$ (launch) or $t = \frac{2 V \sin \alpha}{g}$ .

$R = V \cos \alpha \cdot \frac{2 V \sin \alpha}{g} = \frac{2 V^2 \sin \alpha \cos \alpha}{g} = \frac{V^2 \sin 2 \alpha}{g}$ .

Maximum height: set $\dot y(t) = V \sin \alpha - g t = 0$ , so $t = \frac{V \sin \alpha}{g}$ .

$H = V \sin \alpha \cdot \frac{V \sin \alpha}{g} - \frac{1}{2} g \cdot \left( \frac{V \sin \alpha}{g} \right)^2 = \frac{V^2 \sin^2 \alpha}{g} - \frac{V^2 \sin^2 \alpha}{2 g} = \frac{V^2 \sin^2 \alpha}{2 g}$ .

Markers reward stating the equations of motion, finding times by setting position or velocity components to zero, and the standard expressions for $R$ and $H$ .