Projectile Motion Fundamentals
In August 1971, NASA astronaut David Scott stood on the lunar surface and dropped a 1.32 kg geological hammer and a 0.03 kg falcon feather simultaneously from 1.6 m. With no air resistance, both hit the surface in exactly 1.3 s — confirming Galileo's 1589 prediction that all objects fall at the same rate regardless of mass.
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?
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
Gravity stops acting on a projectile once it reaches its peak.
Air resistance is included in HSC projectile-motion calculations.
Core Content
Resolving the initial velocity into independent 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 back to Earth while moving horizontally at constant speed the entire time. These two motions — vertical (slowed by gravity) and horizontal (constant speed) — run completely independently and share only one thing: time.
Projectile trajectory — launch velocity $v$ resolved into $v_x$ and $v_y$.
Horizontal and vertical motion equations, side by side.
$v_x = v \cos\theta$ — 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:
$v = \sqrt{v_x^2 + v_y^2}$ — magnitude
$\theta = \tan^{-1}\!\left(\dfrac{v_y}{v_x}\right)$ — direction above the horizontal
A ball is kicked at 20 m/s, 30° above the horizontal. Find the horizontal and vertical components of the initial velocity.
- Given. $v = 20 \text{ m/s}$, $\theta = 30°$.
- Find. $v_x$ and $v_y$.
- Method. Use $v_x = v\cos\theta$ and $v_y = v\sin\theta$.
- Solve. $v_x = 20 \cos 30° = 20 \times 0.866 = 17.3 \text{ m/s}$.
- Solve. $v_y = 20 \sin 30° = 20 \times 0.500 = 10.0 \text{ m/s}$.
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 $\theta = \tan^{-1}(v_y/v_x)$ above the horizontal.
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 $v_x$?
Two separate one-dimensional motions sharing only time
We just saw how launch velocity splits into horizontal and vertical components. That raises a question: do these components affect each other during flight? This card answers it → they are completely independent, sharing only time.
The key insight of projectile motion is that horizontal and vertical motion are completely independent. They share only one quantity: time. The time of flight is set entirely by the vertical motion.
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 $a = -g = -9.8 \text{ m/s}^2$:
- $s_y = u_y t + \tfrac{1}{2} a t^2$
- $v_y = u_y + a t$
- $v_y^2 = u_y^2 + 2 a s_y$
Find the time to reach maximum height for a projectile launched at 25 m/s, 40° above the horizontal.
- Given. $v = 25 \text{ m/s}$, $\theta = 40°$, $a = -9.8 \text{ m/s}^2$.
- Find. $t_{\text{up}}$ (time to maximum height).
- Method. At max height, $v_y = 0$. Use $v_y = u_y + at$ with $u_y = v\sin\theta$.
- Solve. $u_y = 25\sin 40° = 25 \times 0.643 = 16.1 \text{ m/s}$.
- Solve. $0 = 16.1 + (-9.8)t \;\Rightarrow\; t = \dfrac{16.1}{9.8} = 1.64 \text{ s}$.
Horizontal ($a_x = 0$, $v_x$ constant, $s_x = v_x t$) and vertical ($a = -9.8\ \text{m/s}^2$, SUVAT equations apply) motions are completely independent; they share only time — the flight time is determined solely by the vertical motion.
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.
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…
Why the path of a projectile 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 $a = -g$:
$s_y = s_x \tan\theta - \dfrac{g \, s_x^2}{2 v^2 \cos^2\theta}$
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.
This derivation is not directly examinable in the HSC, but understanding it deepens your grasp of why the path is parabolic — and helps with band-6 short answers that ask you to explain the shape, not just describe it.
Eliminating time from the displacement equations gives the trajectory equation $s_y = s_x\tan\theta - \dfrac{g\,s_x^2}{2v^2\cos^2\theta}$; the $s_x^2$ term proves the path is a downward-opening parabola (enrichment — not directly examined).
Pause — write the highlighted trajectory equation into your book.
Three of these statements about an ideal projectile are correct. Pick the odd one out.
Fill the gap. For a projectile in symmetric flight, the range formula is $R = \dfrac{v^2 \sin(\_\_\theta)}{g}$. The missing factor is _____.
Launch components: $v_x = v\cos\theta$ · $v_y = v\sin\theta$
Horizontal: $s_x = v_x t$ (constant velocity)
Vertical: $s_y = u_y t + \tfrac{1}{2}at^2$ · $v_y = u_y + at$ · $v_y^2 = u_y^2 + 2as_y$
Constant: $a = -9.8 \text{ m/s}^2$ (upward positive)
Which equation correctly gives vertical displacement for a projectile (upward positive, $a = -g$)?
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 optimise their launch angle around 20° (not 45°) because takeoff speed drops at steeper angles — the body cannot generate maximum horizontal velocity when trying to leap too steeply. It is a classic trade-off between angle and speed.
Why do elite long jumpers take off at about 20°, not 45°?
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?
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 _____.
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?
Misconceptions — final check
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?
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.
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: "A projectile launched at 45° will always travel further than one launched at any other angle from the same speed." Consider the conditions under which this is true, and when it may not apply.
1 mark: identifies the statement as conditional on equal launch and landing height · 1 mark: derivation / quotation of $R = \dfrac{v^2 \sin 2\theta}{g}$ and why $\theta = 45°$ maximises it · 1 mark: explains shallower optimum when launching above landing point · 1 mark: explains steeper optimum when landing above launch (or equivalent insight)
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.
Short Answer — Model Answers
Q1 (2 marks): $v_x = 18 \cos 35° = 18 \times 0.819 = 14.7 \text{ m/s}$ (1 mark). $v_y = 18 \sin 35° = 18 \times 0.574 = 10.3 \text{ m/s}$ (1 mark).
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 true only when launch and landing are at the same height (1 mark). For symmetric flight, the range is $R = \dfrac{v^2 \sin 2\theta}{g}$, which is maximised when $\sin 2\theta = 1$, i.e. when $\theta = 45°$ (1 mark). However, when launched from a height above the landing point, the optimum is less than 45° because the extra fall time rewards a shallower angle that maximises horizontal velocity (1 mark). Conversely, when landing above launch height, the optimum exceeds 45°. So the 45° result is a special case of symmetric flight, not a universal rule (1 mark).
Five timed questions on projectile fundamentals. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaParabolic trajectories — match the projectile to its target. Lighter than the boss — pure aim-and-time practice that hammers home the angle/range relationship.
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 result — both hit the surface in exactly 1.3 s — follows directly from what this lesson covers: vertical motion is governed only by gravitational acceleration ($g_{\text{Moon}} = 1.62 \text{ m/s}^2$), not by the mass of the object. Using $s = \tfrac{1}{2}gt^2$: $t = \sqrt{2 \times 1.6 / 1.62} = 1.40 \text{ s}$. No air resistance meant no drag force to differentiate the 1.32 kg hammer from the 0.03 kg feather — Galileo's 1589 prediction, confirmed 382 years later on another world.
Has your understanding of independent components held up?