Year 12 PhysicsModule 5⏱ ~30 min5 MC · 3 Short AnswerLesson 1 of 18
Projectile Motion Fundamentals
In August 1971, NASA astronaut David Scott dropped a geological hammer and a falcon feather together on the Moon. With negligible atmosphere, both objects shared the same downward acceleration and landed together.
Today's hook: In August 1971, NASA astronaut David Scott dropped a hammer (1.32 kg) and a falcon feather (0.03 kg) together on the Moon's surface. Take an illustrative drop height of 1.6 m: both reach the ground together in about 1.4 s, despite their masses differing by a factor of about 44. (The drop height and time here are illustrative model values, not mission telemetry.) Why does mass make no difference, and what does this tell us about the vertical component of projectile motion?
0/5TASKS
1
You’re here
Orient and choose a frame
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
You throw a ball at the same speed at two different launch angles: 45° and 60° above the horizontal.
Which goes further, and why? Write down your prediction before working through the lesson, you will come back to it at the end.
Warm-up, what single force acts on an ideal projectile (no air resistance) during its entire flight?
Learning Intentions
goals
Know, Resolve Velocity
Resolve launch velocity into horizontal and vertical components using trigonometry
Recombine components to find resultant velocity at any instant
Understand, Apply Equations of Motion
Apply $s = ut + \tfrac{1}{2}at^2$ and $v = u + at$ independently to each direction
Solve for time of flight, maximum height, and range
Can Do, Explain Independence
Explain why horizontal and vertical motions are independent of each other
Justify why only the vertical motion sets the flight time
Model and reference convention
Work in an inertial ground frame with $+x$ horizontal in the launch direction and $+y$ upward. Treat the projectile as a point particle, take gravitational acceleration as uniform and downward ($a_y=-g$, $g=9.8\ \text{m s}^{-2}$ near Earth), neglect air resistance and Earth curvature, and use the same elapsed time $t$ in both component equations.
2
Resolve the launch vector
Scan these before reading
vocab
ProjectileAn object whose motion, in this model, is governed only by gravity — in air with drag neglected, or in a vacuum such as on the Moon.
TrajectoryThe parabolic path traced by a projectile under constant gravitational acceleration.
Time of flightThe total time a projectile remains in the air.
RangeThe total horizontal displacement of a projectile.
Maximum heightThe highest vertical point reached in the trajectory.
ComponentsThe perpendicular parts of a vector, typically horizontal ($v_x$) and vertical ($v_y$).
Cross-lesson links: This module opens by extending Year 11 kinematics into two dimensions. The projectile motion model here (independent horizontal/vertical components) is the foundation for every subsequent topic in M5, circular motion, orbital mechanics, and gravitational fields all depend on resolving motion into components.
Misconceptions to fix
Horizontal motion slows down over time as the projectile flies.
Ignoring air resistance, horizontal velocity is constant because there is no horizontal force, so no horizontal acceleration. It is the vertical velocity that changes, gravity only pulls down.
Gravity only acts at the top of the flight, where the ball "stops".
Gravity acts continuously at $9.8 \text{ m/s}^2$ downward for the entire trajectory, on the way up, at the peak, and on the way down.
At the peak of the flight, acceleration is zero.
Acceleration is always $-9.8 \text{ m/s}^2$. It is the vertical velocity that is momentarily zero at the peak, the acceleration is unchanged.
Gravity stops acting on a projectile once it reaches its peak.
Air resistance is included in HSC projectile-motion calculations.
Core Content
1
Vector Components of Launch Velocity
+5 XP
Resolving the initial velocity into dynamically separable horizontal and vertical parts
Watch a cricket ball leave the bat at 30 m/s at 40° above horizontal: it arcs upward, slows vertically, then falls while moving horizontally at constant speed in the ideal model. The component equations are dynamically separable, but they describe one velocity vector in the same reference frame and share the same elapsed time.
Projectile trajectory, launch velocity $v$ resolved into $v_x$ and $v_y$.
Horizontal and vertical motion equations, side by side.
Component equations
$v_x = v \cos\theta$ , horizontal component (m/s)
$v_y = v \sin\theta$ , vertical component (m/s)
The horizontal component $v_x$ remains constant throughout the flight because there is no horizontal acceleration (ignoring air resistance). The vertical component $v_y$ changes continuously due to gravity acting downward at $9.8 \text{ m/s}^2$.
At any instant during the flight, the resultant velocity is recovered from the components:
Resultant velocity
$v = \sqrt{v_x^2 + v_y^2}$ , magnitude
$\theta = \tan^{-1}\!\left(\dfrac{v_y}{v_x}\right)$ , direction relative to the horizontal
Your calculator's $\tan^{-1}$ only returns angles between $-90^\circ$ and $+90^\circ$, so check the quadrant against the signs of $v_x$ and $v_y$: a negative $v_y$ means the velocity points below the horizontal, and a negative $v_x$ means the motion is to the left, which needs $180^\circ$ added. (This is what the $\operatorname{atan2}(v_y, v_x)$ function does for you.)
Worked example 1+5 XP on full reveal
Finding launch components. A ball is kicked at 20 m/s, 30° above the horizontal. Find the horizontal and vertical components of its initial velocity.
1
Record the known quantities $v = 20 \text{ m/s}$ and $\theta = 30°$.
The speed is the magnitude of the launch vector; the angle is measured above the positive horizontal direction.
2
Select the component equations $v_x = v\cos\theta$ and $v_y = v\sin\theta$.
The horizontal component is adjacent to the launch angle, while the vertical component is opposite it.
At launch the ball is travelling upward, so $v_y$ is positive when upward is the positive direction.
5
State the answer with directions $\mathbf{v_x=17.3\text{ m/s}}$ horizontally and $\mathbf{v_y=10.0\text{ m/s}}$ upward.
Components are vectors. Including their directions makes the sign convention explicit.
A projectile's launch velocity resolves into $v_x = v\cos\theta$ (horizontal, constant throughout flight) and $v_y = v\sin\theta$ (vertical, changes due to gravity); the resultant at any instant is $v = \sqrt{v_x^2 + v_y^2}$ at angle $\theta = \tan^{-1}(v_y/v_x)$ to the horizontal, with the quadrant fixed from the signs of $v_x$ and $v_y$.
Pause, copy the highlighted definition and formula into your book before moving on.
A projectile is launched at 10 m/s, 60° above the horizontal. What is the horizontal component $v_x$?
3
Separate the component motions
2
Independence of Horizontal and Vertical Motion
+5 XP
Two separate one-dimensional motions sharing only time
We just saw how launch velocity splits into horizontal and vertical components. That raises a question: how are the component equations connected? This card answers it → they are dynamically separable under the model while sharing the same elapsed time.
Horizontal and vertical motion can be analysed with separate one-dimensional equations under the ideal model. They remain components of one motion, use the same reference frame and share the same time variable. For a specified vertical displacement, the vertical equation determines the flight time.
Key insight
The time to reach maximum height (and total flight time) depends only on the vertical component of velocity. The horizontal motion runs at constant velocity for exactly that same time.
Horizontal motion, constant velocity
There is no horizontal acceleration ($a_x = 0$):
$s_x = v_x t$ (horizontal displacement)
$v_x$ stays constant throughout
Vertical motion, constant acceleration
Taking upward as positive, with $a = -g = -9.8 \text{ m/s}^2$:
$s_y = u_y t + \tfrac{1}{2} a t^2$
$v_y = u_y + a t$
$v_y^2 = u_y^2 + 2 a s_y$
Worked example 2+5 XP on full reveal
Time to maximum height. Find the time to reach maximum height for a projectile launched at 25 m/s, 40° above the horizontal.
1
Choose a reference convention Take upward as positive, so $a_y=-9.8\text{ m/s}^2$.
Gravity points downward, opposite the chosen positive vertical direction.
2
Resolve the initial vertical velocity $u_y=25\sin40°=25(0.643)=\mathbf{16.1\text{ m/s}}$.
Only the vertical component determines how long the projectile takes to reach its highest point.
3
Use the peak condition At maximum height, the instantaneous vertical velocity is $v_y=0$.
The vertical velocity is momentarily zero at the turning point; the acceleration is still $-9.8\text{ m/s}^2$.
4
Substitute into $v_y=u_y+a_yt$ $0=16.1+(-9.8)t$.
The signs are consistent with the upward-positive reference convention.
5
Solve and state the result $t=\dfrac{16.1}{9.8}=\mathbf{1.64\text{ s}}$.
This is the time to maximum height, not the total flight time.
Under the ideal model, horizontal ($a_x = 0$, $v_x$ constant, $s_x = v_x t$) and vertical ($a_y = -9.8\ \text{m/s}^2$, constant-acceleration equations apply) motions are dynamically separable. They share the same elapsed time; the vertical equation fixes that time for a specified final height.
Add the highlighted principle to your notes before the check below.
The horizontal velocity of a projectile (ignoring air resistance) stays constant throughout the flight.
At the peak of a projectile's trajectory, the acceleration is zero.
The total time of flight is determined entirely by the vertical motion.
Compare launches at 30° and 60°. Keep the launch speed fixed, then observe the range and flight time.
Use the simulator. Two balls leave the ground at the same speed: one at 30°, one at 60°. Compared to each other, the ranges are…
4
Represent the trajectory
Assessment boundary
Core — assessable
Know that an ideal projectile follows a downward-opening parabolic path. Explain this using constant horizontal velocity, uniform downward acceleration, and the shared time variable.
Optional enrichment
Follow how eliminating time produces the trajectory equation. You do not need to memorise or reproduce this derivation for this lesson’s assessment.
3
Optional enrichment, deriving the trajectory equation
Not required to reproduce
Go deeper: why the ideal projectile path is mathematically a parabola
We just saw that horizontal and vertical motions are independent, each described by separate equations. That raises a question: what shape does the combined path trace? This card answers it → eliminating time links the two displacements into a single parabolic equation.
By eliminating time $t$ from the two displacement equations, the path collapses into a single equation $s_y(s_x)$, and that equation is a parabola.
From horizontal motion: $t = \dfrac{s_x}{v_x} = \dfrac{s_x}{v\cos\theta}$.
Substitute into $s_y = u_y t + \tfrac{1}{2} a t^2$, with $u_y = v\sin\theta$ and $a = -g$:
This is a parabola opening downward, the $s_x^2$ term is what makes it parabolic. So the trajectory of any projectile in a uniform gravitational field (neglecting air resistance) is parabolic.
For assessment: be ready to explain the parabolic shape qualitatively from the horizontal and vertical component models. The algebraic elimination of $t$ and the final trajectory equation are extension material here; do not spend revision time memorising this derivation.
Optional note: record the derivation only if it helps you understand the core parabolic-path explanation.
Core checks · assessable content
Three of these statements about an ideal projectile are correct. Pick the odd one out.
Fill the gap. With constant vertical acceleration, displacement is $s_y=u_y t+\dfrac{1}{\_\_}a_y t^2$. The missing denominator is _____.
Core formulae · assessable
Model and frame: point particle, uniform $g$, negligible air resistance, inertial ground frame, $+x$ in the launch direction and $+y$ upward.
Which equation correctly gives vertical displacement for a projectile (upward positive, $a = -g$)?
Core application · model limitation
A long jumper launches at 20° with speed 9.5 m/s. The horizontal component ($v_x = 9.5\cos 20° = 8.9 \text{ m/s}$) carries them forward; the vertical component ($v_y = 9.5\sin 20° = 3.2 \text{ m/s}$) sets the flight time.
Elite jumpers often use angles around 20°, rather than the idealised 45°, because takeoff speed drops at steeper angles and launch and landing heights differ. The best angle depends on the athlete and technique; 20° is contextual, not a universal law.
Why do elite long jumpers take off at about 20°, not 45°?
5
Apply component reasoning
Activity 1, Component Drills
ApplyBand 3
Practise resolving and recombining velocity components
A projectile is launched at 15 m/s, 50° above the horizontal. Find $v_x$ and $v_y$.
At a point in flight, $v_x = 12 \text{ m/s}$ and $v_y = 5 \text{ m/s}$. Find the resultant velocity (magnitude and direction above horizontal).
A ball is thrown horizontally at 8 m/s from a cliff 20 m high. How long until it hits the ground?
Plain-text notation is fine—for example, sqrt(169) or theta = 22.6°.
Drill check, for a projectile launched at 18 m/s, 35° above horizontal, the vertical component $v_y$ (to one decimal place, in m/s) is _____.
Activity 2, Concept Check
UnderstandBand 4
Explain a classic physics demonstration
A coin is launched horizontally off the edge of a table at the same instant a second coin is simply dropped from the same height. Both coins hit the floor at the same time. Explain why, using the concept of independence of horizontal and vertical motion.
During a typical projectile flight (no air resistance), which set correctly describes what is constant and what is changing?
6
Consolidate the model and its limits
Wrap-up, Misconceptions & Summary
Misconceptions, final check
"The 60° launch always goes further than the 30° launch."
For equal launch and landing height, the range depends on $\sin 2\theta$. Since $\sin 60° = \sin 120°$, the 30° and 60° launches give exactly the same range at the same speed. The maximum range is at 45°.
"I can use $s = v_x t$ to find the time of flight from the horizontal motion."
The horizontal equation alone has two unknowns ($s_x$ and $t$). Time of flight always comes from the vertical equations, they are the ones tied to gravity.
Copy into your books
Key Definitions
Projectile: object moving under gravity alone
Trajectory: the parabolic path traced
Time of flight: total time airborne
Component Formulae
$v_x = v\cos\theta$
$v_y = v\sin\theta$
$v = \sqrt{v_x^2 + v_y^2}$
Equations of Motion
$s = ut + \tfrac{1}{2}at^2$
$v = u + at$
$v^2 = u^2 + 2as$
Key Principles
Horizontal: $a=0$, $v_x$ constant
Vertical: $a = -9.8 \text{ m/s}^2$
Only vertical motion sets the time
Same launch speed, flat ground. Which two angles give the same range?
✓
Independent practice
Quick recall, projectile fundamentals
+5 XP
A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
Short Answer, 9 marks
+5 XP
ApplyBand 4(2 marks) 1. A projectile is launched at 18 m/s at 35° above the horizontal. Calculate the horizontal and vertical components of the initial velocity.
1 mark: correct $v_x$ with working · 1 mark: correct $v_y$ with working
ApplyBand 5(3 marks) 2. A ball is thrown horizontally at 12 m/s from a cliff 30 m above sea level. Calculate (a) the time to reach the water, (b) the horizontal distance travelled.
1 mark: correctly identifying $u_y = 0$ and using $s_y = \tfrac{1}{2}gt^2$ · 1 mark: correct $t$ · 1 mark: correct horizontal distance using $s_x = v_x t$
EvaluateBand 6(4 marks) 3. Evaluate the statement: "The horizontal and vertical motions of a projectile are completely independent." Explain what is separable, what remains shared, and the model conditions required.
1 mark: horizontal and vertical equations can be solved separately under the ideal model · 1 mark: both components describe one motion and share the same elapsed time/reference frame · 1 mark: horizontal acceleration is zero while vertical acceleration is $-g$ · 1 mark: states relevant model conditions such as negligible drag and uniform gravity
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Q2 (3 marks): (a) Thrown horizontally so $u_y = 0$. Using $s_y = \tfrac{1}{2} g t^2$ with $s_y = 30 \text{ m}$ and $g = 9.8 \text{ m/s}^2$: $30 = 4.9 t^2$, so $t = \sqrt{30/4.9} = 2.47 \text{ s}$ (2 marks, method + answer). (b) $s_x = v_x \, t = 12 \times 2.47 = 29.7 \text{ m}$ (1 mark).
Q3 (4 marks): The statement is only partly correct. Under the ideal projectile model, the horizontal and vertical component equations are dynamically separable, so each direction can be analysed with its own one-dimensional equation (1 mark). They still describe one projectile in the same inertial ground frame and share the same elapsed time $t$ (1 mark). With negligible drag, $a_x=0$ while $a_y=-g$, where $g$ is treated as uniform and downward (1 mark). Therefore “completely independent” is too strong: the separation is a modelling method that assumes a point particle, negligible air resistance, uniform gravity and distances small enough to neglect Earth curvature (1 mark).
✓
Retrieve and reflect
Check what actually stuck
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Parabolic trajectories, match the projectile to its target. Lighter than the boss, pure aim-and-time practice that hammers home the angle/range relationship.
How did your thinking change?
At the start you were asked about the August 1971 Apollo 15 experiment in which NASA astronaut David Scott dropped a hammer and a feather simultaneously from 1.6 m on the Moon.
The shared landing time follows from the model: both objects began with the same vertical velocity and experienced the same lunar gravitational acceleration ($g_{\text{Moon}} \approx 1.62\ \text{m s}^{-2}$). For an illustrative 1.6 m drop, $t = \sqrt{2s/g} = \sqrt{3.2/1.62} \approx 1.4\ \text{s}$. With negligible atmosphere, there was no appreciable drag difference to separate their motion.
Has your understanding of independent components held up?
