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.
The range of a projectile launched at angle $\theta$ with initial speed $v$ is $R = \dfrac{v^2 \sin 2\theta}{g}$. Without calculating — at what value of $\theta$ is $R$ maximised? Write your reasoning.
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.
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: $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$
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
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)
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:
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 $R_{\max} = \dfrac{v^2}{g}$.
$R = \dfrac{20^2 \sin(2 \times 30°)}{10} = \dfrac{400 \sin 60°}{10} = \dfrac{400 \times \dfrac{\sqrt{3}}{2}}{10} = \dfrac{200\sqrt{3}}{10} = 20\sqrt{3} \approx 34.6 \text{ m}$
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?
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:
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 $t = 0$).
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
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 $g = 10 \text{ m/s}^2$).
So $\theta \approx 31.3°$ or $\theta \approx 58.7°$.
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 $x = 2.5$.
First positive solution: $2t = \dfrac{\pi}{3} + \dfrac{\pi}{4} = \dfrac{4\pi+3\pi}{12} = \dfrac{7\pi}{12}$, so $t = \dfrac{7\pi}{24} \approx 0.916$ s.
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 $x = 3$.
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
Did you get this? True or false: A projectile launched at $\theta = 120°$ has the same range as one launched at $\theta = 60°$.
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?
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 $45°$?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
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)
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)
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].
Five timed contextual trig questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering contextual trig questions. Lighter alternative to the boss.
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