Learning Intention 1

Resolve Velocity

Resolve launch velocity into horizontal and vertical components using trigonometry
Recombine components to find resultant velocity at any instant

Learning Intention 2

Apply Equations of Motion

Apply $s = ut + \\frac{1}{2}at^2$ and $v = u + at$ independently to each direction
Solve for time of flight, maximum height, and range

Learning Intention 3

Explain Independence

Explain why horizontal and vertical motions are independent
Understand that only vertical motion determines flight time

Think First — Predict

If you throw a ball at 45° vs 60° from horizontal, which goes further? Why do you think so?

Year 12 Physics Module 5: Advanced Mechanics 45 min Lesson 1 of 18 IQ1: Projectile Motion

Projectile Motion Fundamentals

Learn to analyse the motion of objects projected through the air. Resolve velocity into horizontal and vertical components, apply equations of motion independently in each direction, and understand why the trajectory is parabolic.

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Core Concepts

Vector Components of Launch Velocity

Resolving the initial velocity into independent horizontal and vertical parts

Projectile Trajectory

Projectile Fundamentals

Any projectile launched with speed $v$ at angle $\theta$ above the horizontal has two independent components of initial velocity:

Component Equations

$v_x = v \\cos\\theta$ $v_x$ = horizontal component (m/s)

$v_y = v \\sin\\theta$ $v_y$ = 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 can be found from the components:

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Resultant Velocity

$v = \\sqrt{v_x^2 + v_y^2}$ magnitude of resultant velocity

$\\theta = \\tan^{-1}\\left(\\dfrac{v_y}{v_x}\\right)$ direction above the horizontal

Worked Example — Finding Components

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 $v_y = v \\sin\\theta$

ANS

$v_x = 20 \\times \\cos 30° = 20 \\times 0.866 = 17.3 \\text{ m/s}$

$v_y = 20 \\times \\sin 30° = 20 \\times 0.500 = 10.0 \\text{ m/s}$

Independence of Horizontal and Vertical Motion

Two separate one-dimensional motions sharing only time

The key insight of projectile motion is that horizontal and vertical motions are completely independent. They share only one quantity: time. The time of flight is determined entirely by the vertical motion.

Key Insight

The time to reach maximum height (and total flight time) depends only on the vertical component of velocity. The horizontal motion continues at constant velocity for exactly that same time.

Horizontal Motion (constant velocity)

Since 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 $a = -g = -9.8 \\text{ m/s}^2$:

$s_y = u_y t + \\frac{1}{2} a t^2$
$v_y = u_y + a t$
$v_y^2 = u_y^2 + 2 a s_y$

Worked Example — Time to Maximum Height

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 $u_y = v \\sin\\theta$.

ANS

$u_y = 25 \\times \\sin 40° = 25 \\times 0.643 = 16.1 \\text{ m/s}$

$0 = 16.1 + (-9.8)t$

$t = \\dfrac{16.1}{9.8} = 1.64 \\text{ s}$

The Trajectory Equation

Enrichment — deriving the path of a projectile

By eliminating time $t$ from the horizontal and vertical displacement equations, we obtain the equation of the parabolic path:

From horizontal motion: $t = \\dfrac{s_x}{v_x} = \\dfrac{s_x}{v \\cos\\theta}$

Substitute into $s_y = u_y t + \\frac{1}{2} a t^2$:

$s_y = s_x \\tan\\theta - \\dfrac{g \\cdot s_x^2}{2 v^2 \\cos^2\\theta}$

This is the equation of a parabola opening downward, which confirms that the trajectory of any projectile (in a uniform gravitational field, neglecting air resistance) is parabolic.

Enrichment

This derivation is not examinable in the HSC, but understanding it deepens your grasp of why the path is parabolic. The $s_x^2$ term is what makes it a parabola.

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Essential Formulae — Projectile Motion

$v_x = v \\cos\\theta$ horizontal component of launch velocity

$v_y = v \\sin\\theta$ vertical component of launch velocity

$s_x = v_x t$ horizontal displacement (constant velocity)

$s_y = u_y t + \\frac{1}{2} a t^2$ vertical displacement ($a = -9.8 \\text{ m/s}^2$)

$v_y = u_y + a t$ vertical velocity at time $t$

$v_y^2 = u_y^2 + 2 a s_y$ velocity-displacement relation (no time needed)

Common Misconceptions

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"Horizontal motion slows down over time" — No. Horizontal velocity is constant because there is no horizontal acceleration (ignoring air resistance).

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"Gravity only acts at the top of the flight" — No. Gravity acts constantly at $9.8 \\text{ m/s}^2$ downward throughout the entire trajectory.

✗"The acceleration is zero at the peak" — No. Acceleration is always $-9.8 \\text{ m/s}^2$. It is the vertical velocity that is zero at the peak, not the acceleration.

Real World — Long Jump

Optimising Launch Angle in Athletics

A long jumper launches at 20° with speed 9.5 m/s. The horizontal component ($v_x = 9.5 \\times \\cos 20° = 8.9 \\text{ m/s}$) carries them forward while the vertical component ($v_y = 9.5 \\times \\sin 20° = 3.2 \\text{ m/s}$) determines flight time.

Elite jumpers optimise their launch angle around 20° (not 45°) because takeoff speed decreases at steeper angles. The body cannot generate maximum horizontal velocity when trying to jump too steeply. This is a classic trade-off between angle and speed.

Activities

Activity

Component Drills

practise resolving and recombining velocity components

1 A projectile is launched at 15 m/s, 50° above the horizontal. Find $v_x$ and $v_y$.

2 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).

3 A ball is thrown horizontally at 8 m/s from a cliff 20 m high. How long until it hits the ground?

Concept Check

Independence of Motion

explain a classic physics demonstration

Explain why a projectile launched horizontally and one dropped from the same height hit the ground simultaneously. Use the concept of independence of horizontal and vertical motion in your answer.

📝 Copy into Books

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Key Definitions

Projectile: an object moving under gravity alone
Trajectory: the path of a projectile
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

$s = ut + \\frac{1}{2}at^2$
$v = u + at$
$v^2 = u^2 + 2as$

Key Principles

Horizontal: $a = 0$, $v_x$ constant
Vertical: $a = -9.8 \\text{ m/s}^2$
Only vertical motion determines time

Revisit Your Prediction

Now that you have studied the fundamentals, revisit your earlier prediction about 45° vs 60°.

For the same launch speed, the range depends on $\\sin(2\\theta)$. Since $\\sin(90°) = 1$ (maximum), 45° gives the greatest range when launch and landing are at the same height. At 60°, the range is the same as at 30° because $\\sin(120°) = \\sin(60°)$.

Has your understanding changed? Write a revised explanation:

Interactive: Projectile Launcher Interactive

Interactive: Projectile Launcher Simulator

Inquiry Question 1: Projectile Motion

Key Terms

Projectile An object moving through the air under the influence of gravity alone

Trajectory The parabolic path traced by a projectile through the air

Time of flight The total time a projectile remains in the air

Range The total horizontal displacement of a projectile

Maximum height The highest vertical point reached in the trajectory

Components The perpendicular parts of a vector (horizontal and vertical)

Multiple Choice

1. A ball is thrown at 30° to the horizontal. Which statement is correct?

A Both horizontal and vertical velocities are constant

B Horizontal velocity is constant and vertical acceleration is constant

C Horizontal acceleration is 9.8 m/s²

D Vertical velocity is constant

2. A projectile is launched at 20 m/s, 60° above horizontal. What is $v_x$?

A 20 m/s

B 17.3 m/s

C 10 m/s

D 0 m/s

3. At the peak of a projectile's trajectory:

A Both velocity and acceleration are zero

B Velocity is zero and acceleration is 9.8 m/s² down

C Vertical velocity is zero and acceleration is 9.8 m/s² down

D Acceleration is zero and vertical velocity is 9.8 m/s²

4. A ball is thrown horizontally from a height of 45 m at 10 m/s. How far horizontally does it travel?

A 10 m

B 30 m

C 45 m

D 90 m

5. For a projectile launched from ground level, the time to reach maximum height is:

A Equal to the total flight time

B Half the total flight time

C Twice the total flight time

D Independent of launch angle

Short Answer

Apply Band 4 2 marks

A projectile is launched at 18 m/s at 35° above the horizontal. Calculate the horizontal and vertical components of the initial velocity.

Apply Band 5 3 marks

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.

Evaluate Band 6 4 marks

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.

Model Answers

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Question 1 (2 marks)

$v_x = 18 \\cos 35° = 14.7 \\text{ m/s}$ (1 mark)

$v_y = 18 \\sin 35° = 10.3 \\text{ m/s}$ (1 mark)

Question 2 (3 marks)

(a) $s_y = \\frac{1}{2}gt^2$ since $u_y = 0$ (thrown horizontally)

$30 = 4.9 \\times t^2$

$t = \\sqrt{\\frac{30}{4.9}} = 2.47 \\text{ s}$ (2 marks for method + answer)

(b) $s_x = v_x \\times t = 12 \\times 2.47 = 29.7 \\text{ m}$ (1 mark)

Question 3 (4 marks)

The statement is true only when launch and landing are at the same height (1 mark). In this case, the range $R = \\frac{v^2 \\sin(2\\theta)}{g}$ is maximised when $\\sin(2\\theta) = 1$, which occurs at $\\theta = 45°$ (1 mark).

However, when launched from a height above the landing point, the optimal angle is less than 45° because the extra fall time benefits a shallower angle that maximises horizontal velocity (1 mark). Conversely, when landing above launch height, the optimal angle is greater than 45°. The 45° result assumes symmetrical flight where $s_y = 0$ at both start and end (1 mark).

Boss Battle

✓

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