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Module 7 · L14 of 20 ~45 min ⚡ +110 XP available

Contextual Trigonometric Problems

A ball on a spring oscillates as $x(t) = A\cos(\omega t + \phi)$. A projectile traces a parabola whose range depends on $\sin 2\theta$. Trigonometry is everywhere in motion, and in this lesson you'll learn the four-step method that converts messy context into clean equations, solve them, and translate the answer back into real-world meaning.

Today's hook, A projectile is launched with speed $v$ at angle $\theta$. Its range is $R = \dfrac{v^2 \sin 2\theta}{g}$. Without calculating, at what launch angle is the range maximised? Write your gut answer, you'll verify it after card 05.

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Recall, your gut answer first

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The range of a projectile launched at angle $\theta$ with initial speed $v$ is $R = \dfrac{v^2 \sin 2\theta}{g}$. Without calculatingat what value of $\theta$ is $R$ maximised? Write your reasoning.

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The four-step method

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All contextual trig problems, whether projectile motion, harmonic oscillation, or geometry, follow the same four steps:

Step 1, Draw a diagram. Sketch the physical situation. Label all given quantities and mark what you need to find.

Step 2, Identify the trig relationship. Recognise which formula applies (range formula, SHM equation, auxiliary angle, etc.).

Step 3, Set up and solve. Write the equation, substitute known values, and solve using appropriate trig techniques.

Step 4, Interpret. Translate your mathematical answer back into the real-world context. Check units, domain, and reasonableness.

diagram $\to$ model $\to$ solve $\to$ interpret

Diagram first, always

Drawing a clear diagram often reveals the trig relationship immediately. It also earns marks in the HSC, a labelled diagram is evidence of understanding.

Units matter

Angles can be in degrees or radians. Make sure your calculator mode matches the problem. Time is usually in seconds, distance in metres, check everything.

Interpret back

A mathematical answer of $\theta = -0.3$ rad might be valid, but interpret it: is a negative angle physically meaningful? Always check against the context.

What you'll master

Know

Key facts

Projectile range formula: $R = \dfrac{v^2 \sin 2\theta}{g}$, maximised at $\theta = 45°$
Simple harmonic motion: $x = A\cos(\omega t + \phi)$, period $T = \dfrac{2\pi}{\omega}$
Auxiliary angle form: $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$

Understand

Concepts

How the trig function parameters (amplitude, period, phase) connect to physical quantities
Why the maximum of $\sin 2\theta$ at $2\theta = 90°$ gives $\theta = 45°$ for maximum range
How contextual constraints restrict the domain of valid solutions

Can do

Skills

Apply the four-step method to projectile and harmonic motion problems
Set up and solve trig equations arising from real contexts
Interpret solutions in terms of the physical situation (including rejecting non-physical answers)

Key terms

Projectile motionMotion under gravity only. Range is $R = \dfrac{v^2\sin 2\theta}{g}$ where $\theta$ is the angle of projection and $v$ is the launch speed.

Simple harmonic motion (SHM)Oscillating motion where the restoring force is proportional to displacement. Modelled by $x = A\cos(\omega t + \phi)$ or $x = A\sin(\omega t + \phi)$.

Amplitude $A$The maximum displacement from the equilibrium position in an SHM model.

Angular frequency $\omega$The rate of oscillation in radians per second. Period $T = 2\pi/\omega$, frequency $f = \omega/(2\pi)$.

Phase shift $\phi$A horizontal shift of the trig function, determined by the initial conditions of the system.

Elevation angle $\theta$The angle above the horizontal at which a projectile is launched. Physical constraints require $0° \leq \theta \leq 90°$.

Projectile motion, the range formula

core concept

When a projectile is launched from ground level with speed $v \text{ m/s}$ at angle $\theta$ above the horizontal, and gravity $g \text{ m/s}^2$ acts downward, the horizontal range is:

$$R = \frac{v^2 \sin 2\theta}{g}$$

Maximising range: $R$ is maximised when $\sin 2\theta$ is maximised, i.e. when $\sin 2\theta = 1$. This gives $2\theta = 90°$, so $\theta = 45°$. The maximum range is $R_{\max} = \dfrac{v^2}{g}$.

Example. A ball is kicked at $20 \text{ m/s}$ at $30°$ to the horizontal. Find the range (take $g = 10 \text{ m/s}^2$).
$R = \dfrac{20^2 \sin(2 \times 30°)}{10} = \dfrac{400 \sin 60°}{10} = \dfrac{400 \times \dfrac{\sqrt{3}}{2}}{10} = \dfrac{200\sqrt{3}}{10} = 20\sqrt{3} \approx 34.6 \text{ m}$

Equal-range angles: Since $\sin 2\theta = \sin(180° - 2\theta) = \sin 2(90° - \theta)$, angles $\theta$ and $90° - \theta$ give the same range. For example, $30°$ and $60°$ produce identical ranges.

Range formula: R = v^2 2{g}, max when = 45°, giving R_{} = v^2{g}; Complementary angles give equal range: and 90° -

Pause, copy the range formula into your book: $R = \frac{v^2\sin 2\theta}{g}$; maximum range $R_{\max} = \frac{v^2}{g}$ at $\theta = 45°$; complementary angles $\theta$ and $90°-\theta$ give the same range.

Quick check: A projectile launched at $60°$ has the same range as one launched at which other angle?

Simple harmonic motion

core concept

We just saw that projectile range $R = \frac{v^2 \sin 2\theta}{g}$ is maximised at $45°$ and that complementary angles give equal range. That raises a question: oscillating systems, springs, pendulums, also use trig functions; how is the SHM displacement model written and what do its parameters mean? This card answers it → $x = A\sin(\omega t + \phi)$ where $A$ is amplitude, $\omega$ is angular frequency, $T = 2\pi/\omega$ is the period.

A particle undergoing SHM has displacement from equilibrium modelled by:

$$x(t) = A\cos(\omega t + \phi) \quad \text{or} \quad x(t) = A\sin(\omega t + \phi)$$

Key relationships:

Period: $T = \dfrac{2\pi}{\omega}$, the time for one complete oscillation.
Frequency: $f = \dfrac{1}{T} = \dfrac{\omega}{2\pi}$, oscillations per second (Hz).
Amplitude: $A$, the maximum displacement.
Phase: $\phi$, determined by the initial conditions (position and velocity at $t = 0$).

Finding the model from initial conditions. Suppose $x(0) = 3$ m and $\dot{x}(0) = 0$ (particle starts at rest at maximum displacement). Then $x = A\cos(\omega t + \phi)$ with $\phi = 0$ and $A = 3$: $x(t) = 3\cos(\omega t)$.

Solving for time: To find when the particle first reaches $x = c$ (where $|c| \leq A$), solve $\cos(\omega t + \phi) = \dfrac{c}{A}$ for the smallest positive $t$.

SHM model: x = A( t + ) where A = amplitude, = angular frequency, = phase; Period: T = 2/; Frequency: f = /(2)

Pause, copy the SHM model into your book: $x = A\sin(\omega t + \phi)$ with amplitude $A$, angular frequency $\omega$, phase $\phi$; period $T = 2\pi/\omega$ and frequency $f = \omega/(2\pi)$.

Did you get this? True or false: A particle in SHM described by $x = 4\cos(3t)$ has period $T = \dfrac{2\pi}{3}$ seconds.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · PROJECTILE RANGE

A stone is thrown at $15 \text{ m/s}$. Find the two possible angles of projection that give a range of $20 \text{ m}$ (use $g = 10 \text{ m/s}^2$).

Draw a diagram. Apply the range formula: $R = \dfrac{v^2 \sin 2\theta}{g}$.

Identify: $v = 15$, $R = 20$, $g = 10$, unknown $\theta$.

PROBLEM 2 · HARMONIC MOTION

A particle oscillates with position $x = 5\cos\!\left(2t - \dfrac{\pi}{4}\right)$ metres at time $t$ seconds. Find: (a) the amplitude; (b) the period; (c) the first time $t > 0$ when $x = 2.5$.

(a) Amplitude $= A = 5$ m. (b) Period $= T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{2} = \pi$ seconds.

Read off directly: $A = 5$ (coefficient), $\omega = 2$ (coefficient of $t$).

PROBLEM 3 · GEOMETRY IN CONTEXT

A ladder of length 5 m leans against a wall. The foot is $x$ metres from the wall. The top makes angle $\alpha$ with the wall. Express $x$ in terms of $\alpha$ and find the value of $\alpha$ (to the nearest degree) when $x = 3$.

Draw a right-angled triangle. The ladder is the hypotenuse (5 m), $x$ is the opposite side to angle $\alpha$, the wall is the adjacent side.

Label the diagram carefully, $\alpha$ is the angle at the top of the ladder (between ladder and wall).

Fill the gap: For the SHM model $x = 3\cos(4t)$, the period is $T = \dfrac{2\pi}{4} = $ seconds.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the double angle in the range formula

The range formula uses $\sin 2\theta$, not $\sin\theta$. Writing $R = \dfrac{v^2\sin\theta}{g}$ is wrong. The double angle arises because horizontal and vertical components both contribute: $R = \dfrac{2v^2\sin\theta\cos\theta}{g} = \dfrac{v^2\sin 2\theta}{g}$.

Trap 02

Confusing amplitude and period in SHM

In $x = A\cos(\omega t + \phi)$: $A$ is the amplitude (maximum displacement), NOT the period. The period is $T = 2\pi/\omega$. Students often write $T = A$ or confuse the multiplier $\omega$ with the period directly. Always use $T = 2\pi/\omega$.

Trap 03

Not checking domain for physical validity

Solving a trig equation may give multiple solutions, but only some are physically valid. For a projectile, $\theta \in [0°, 90°]$, angles outside this range have no meaning. For SHM, we need $t \geq 0$. Always filter solutions through the physical constraints.

Did you get this? True or false: A projectile launched at $\theta = 120°$ has the same range as one launched at $\theta = 60°$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A ball is thrown at $25 \text{ m/s}$. Using $g = 10 \text{ m/s}^2$, find the maximum horizontal range and the angle at which it occurs.

A particle in SHM has position $x = 6\sin(2\pi t)$ metres. Find: (a) the amplitude; (b) the period; (c) the first positive $t$ when $x = 3$.

A projectile launched at $20 \text{ m/s}$ achieves a range of $30 \text{ m}$. Find the two possible launch angles, correct to the nearest degree ($g = 10$).

A particle starts at $x = 0$ with velocity $6 \text{ m/s}$ in SHM with angular frequency $\omega = 3$. Write an expression for $x(t)$ and find the amplitude.

Explain why the range formula $R = \dfrac{v^2\sin 2\theta}{g}$ gives equal ranges for $\theta = 35°$ and $\theta = 55°$. What is the angle that maximises the range?

Odd one out: Three of these statements are correct. Which one is WRONG?

Revisit your thinking

Earlier you estimated the launch angle that maximises the range $R = \dfrac{v^2\sin 2\theta}{g}$.

The answer is $\theta = 45°$. Since $R \propto \sin 2\theta$, we maximise $\sin 2\theta$. The sine function reaches its maximum of 1 when $2\theta = 90°$, so $\theta = 45°$. Interestingly, for any target range less than the maximum, there are exactly two valid launch angles, one shallower and one steeper than $45°$, summing to $90°$. Did your intuition identify $45°$?

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Multiple choice

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Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A ball is thrown at $30 \text{ m/s}$ at $45°$ to the horizontal. Using $g = 10 \text{ m/s}^2$, find its range. (2 marks)

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ApplyBand 43 marks

Q2. A particle oscillates with $x = 8\cos\!\left(\dfrac{\pi t}{2}\right)$ metres. Find the amplitude, period, and the first positive time when $x = 4$. (3 marks)

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AnalyseBand 53 marks

Q3. A projectile is launched at speed $v$ over level ground. Show that if the range equals the maximum possible range divided by $\sqrt{2}$, then the two valid launch angles sum to $90°$ and are each $22.5°$ from $45°$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:
1. $R_{\max} = v^2/g = 625/10 = 62.5$ m at $\theta = 45°$.
2. (a) $A = 6$ m. (b) $T = 2\pi/(2\pi) = 1$ s. (c) $6\sin(2\pi t) = 3 \Rightarrow \sin(2\pi t) = 0.5 \Rightarrow 2\pi t = \pi/6 \Rightarrow t = 1/12$ s.
3. $\sin 2\theta = 30 \times 10/400 = 0.75 \Rightarrow 2\theta = 48.6°$ or $131.4° \Rightarrow \theta \approx 24°$ or $66°$.
4. $x(0)=0$ and $\dot{x}(0) = 6$: use $x = A\sin(3t)$; $\dot{x} = 3A\cos(3t)$, at $t=0$: $3A = 6 \Rightarrow A = 2$ m.
5. $\sin 2(35°) = \sin 70°$; $\sin 2(55°) = \sin 110° = \sin 70°$ (since $\sin(180°-x) = \sin x$). Max range at $\theta = 45°$.

Q1 (2 marks): $R = \dfrac{30^2\sin(90°)}{10} = \dfrac{900 \times 1}{10} = \mathbf{90}$ m [1 for formula, 1 for answer].

Q2 (3 marks): $A = 8$ m [1]; $\omega = \pi/2$, $T = 2\pi/(\pi/2) = 4$ s [1]; $8\cos(\pi t/2) = 4 \Rightarrow \cos(\pi t/2) = 1/2 \Rightarrow \pi t/2 = \pi/3 \Rightarrow t = 2/3$ s [1].

Q3 (3 marks): $R_{\max} = v^2/g$ [1]. Setting $R = R_{\max}/\sqrt{2}$: $\sin 2\theta = 1/\sqrt{2}$ [1]. Solutions: $2\theta = 45°$ or $2\theta = 135°$, giving $\theta = 22.5°$ or $\theta = 67.5°$. Sum = $90°$; each is $45° - 22.5° = 22.5°$ from $45°$ ✓ [1].

Boss battle · The Context Solver

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Five timed contextual trig questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering contextual trig questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 13 · Graphical Solutions Lesson 15 · Maximising and Minimising Trig Expressions →

Module overview · Extension 1 · Checkpoint 3