Year 12 Physics Module 5 ⏱ ~30 min 5 MC · 3 Short Answer Lesson 3 of 18

Problem Solving: Projectiles

Turn a crowded projectile question into a reliable sequence: choose a reference frame, list signed quantities, select equations whose conditions are satisfied, solve one component at a time, then report a physically complete answer.

Today's hook: A launch speed such as $18\ \text{m/s}$ is a vector magnitude, not either component. By the end, you will know how to translate every value into a signed horizontal or vertical quantity before choosing an equation.

0/5TASKS

You’re here

Choose the frame and decode the question

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response

Before You Read

warm-up

Write down every formula you know for projectile motion. Include what each variable means and its units.

Learning Intentions

goals

Know

The GIVEN → FIND → METHOD → ANSWER problem-solving protocol
The six key projectile motion formulas and when each applies
When the range equation $R = v^2\sin(2\theta)/g$ can and cannot be used

Understand

Why launch height vs landing height determines which equations apply
How to solve problems with partial information by combining vertical and horizontal equations
Why time of flight depends on vertical motion only, not horizontal velocity

Can Do

Apply the problem-solving protocol to multi-step projectile questions
Identify which equation to use based on known and unknown quantities
Solve projectile problems involving uneven ground or intermediate obstacles

Scan these before reading

vocab

Range ($R$)Horizontal distance from launch to landing.

Time of flightTotal elapsed time from launch to landing.

Maximum heightPeak vertical position above the launch point.

Components$v_x = v\cos\theta$, $v_y = v\sin\theta$, always resolved at launch.

Symmetrical flightLaunch and landing at the same height; total flight time $= 2t_{\text{up}}$.

Problem-solving protocolGIVEN → FIND → METHOD → ANSWER, a structured approach to every question.

Cross-lesson links: L02 derived the equations. L03 applies them systematically, the GIVEN→FIND→METHOD→ANSWER protocol for multi-step projectile problems is exactly the exam technique you need, whether the object is a tennis ball or a Mars lander.

Misconceptions To Fix

watch out

✗ Wrong: Use the range equation $R = v^2\sin(2\theta)/g$ for all projectile problems.

✓ Right: The range equation only applies on level ground. Always check if launch and landing heights differ. If they do, solve using the equations of motion separately.

✗ Wrong: Time of flight depends on horizontal velocity.

✓ Right: Time is determined entirely by vertical motion. The horizontal velocity $v_x$ affects range but not flight time (ignoring air resistance).

✗ Wrong: You need to know the mass of the projectile.

✓ Right: Projectile motion neglecting air resistance is completely independent of mass. All objects follow the same trajectory for the same launch conditions.

Core Content

The Problem-Solving Protocol

+5 XP

A structured approach to every projectile question

Projectile Worked Example

Problem Solving Projectiles

Multi-step projectile problems can seem overwhelming. The key is to follow a consistent protocol that breaks the problem into manageable steps.

GIVEN → FIND → METHOD → ANSWER

GIVEN: List all known quantities with units and signs. Resolve the launch velocity into horizontal and vertical components immediately: $v_x = v\cos\theta$ and $v_y = v\sin\theta$.

FIND: State clearly what you need to calculate.

METHOD: Write the equation(s) you will use, rearranged for the unknown. Name the principle (e.g., “equation of motion in vertical direction”).

ANSWER: Substitute values with units, calculate, and check reasonableness.

FRAME FIRST

Use an inertial ground frame. Put the origin at launch unless another origin makes the comparison clearer. Let $+x$ follow the launch direction and $+y$ point upward, so $a_x=0$ and $a_y=-9.8\ \text{m/s}^2$ near Earth.

A stated launch speed $v$ is the magnitude of the initial velocity. It becomes $u_x=v\cos\theta$ and $u_y=v\sin\theta$; do not substitute the total speed as either component.

Before calculating anything, ask: Are launch and landing at the same vertical position? The shortcut $R=v^2\sin(2\theta)/g$ is valid only for a point projectile in uniform gravity with negligible air resistance, equal launch and landing heights, and angle measured from the horizontal. Otherwise solve the component equations.

Problem-Solving Checklist

Choose the origin, reference frame and positive directions
Draw a diagram and list every known quantity with its unit and sign
Resolve the launch velocity: $u_x=v\cos\theta$, $u_y=v\sin\theta$
Check model and height conditions before using a shortcut
Choose the right equation, match knowns/unknowns
Substitute, calculate, check (units, sign, magnitude)

FRAME → GIVEN → FIND → METHOD → ANSWER. Choose $+x$ and $+y$ first; then attach signs and units. Use $R=v^2\sin(2\theta)/g$ only for equal-height flight under uniform $g$ with negligible drag.

Pause, copy the highlighted protocol into your book before moving on.

Did you get this? True or false: the range equation $R = v^2\sin(2\theta)/g$ can be used whenever a projectile is launched at an angle.

You’re here

Solve an equal-height problem

Worked Example 1: Finding Launch Speed from Range

+5 XP

Using the range equation on level ground

We just saw the four-step protocol for tackling projectile problems. That raises a question: how does the protocol work in practice when the range equation can be used? This card answers it → worked example on level ground, rearranging $R = v^2\sin(2\theta)/g$ for launch speed.

PROBLEM 1 · RANGE EQUATION

A projectile is launched at 38° above the horizontal and lands 45 m away on level ground. Find the launch speed. Take $g = 9.8\ \text{m/s}^2$.

GIVEN: $R = 45\ \text{m}$, $\theta = 38°$, level ground, $g = 9.8\ \text{m/s}^2$
FIND: Launch speed $v$

Identify all given quantities and state the unknown before choosing an equation.

To find launch speed from range on level ground, rearrange the range equation: $v = \sqrt{Rg/\sin(2\theta)}$. Always compute $\sin(2\theta)$ first, and confirm level-ground conditions before applying the formula.

Add the highlighted rearrangement to your notes before the check below.

Quick check: A ball is launched at 45° and lands 80 m away on level ground. Which value of $\sin(2\theta)$ is used in the range equation?

You’re here

Connect height, time and range

Worked Example 2: Maximum Height, Time and Range

+5 XP

A complete analysis from launch to landing

We just saw how to find launch speed using the range equation. That raises a question: what if the question asks for maximum height, time, and range all together? This card answers it → use $h_{\text{max}} = v_y^2/(2g)$, $t_{\text{up}} = v_y/g$, $t_{\text{total}} = 2t_{\text{up}}$, and $R = v_x t_{\text{total}}$ in sequence.

PROBLEM 2 · COMPLETE TRAJECTORY

A ball is thrown at $25\ \text{m/s}$ at $50°$ above the horizontal from ground level. Find: (a) the maximum height, (b) the time to reach maximum height, (c) the total flight time, and (d) the horizontal range. Take $g = 9.8\ \text{m/s}^2$.

GIVEN & components:
$v = 25\ \text{m/s}$, $\theta = 50°$, level ground, $g = 9.8\ \text{m/s}^2$
$v_x = 25\cos(50°) = 16.1\ \text{m/s}$, $\quad v_y = 25\sin(50°) = 19.2\ \text{m/s}$

Resolve components first, this is step 2 of the checklist and enables all subsequent calculations.

For symmetrical flight on level ground: $h_{\text{max}} = v_y^2/(2g)$; $t_{\text{up}} = v_y/g$; $t_{\text{total}} = 2t_{\text{up}}$; range $R = v_x \times t_{\text{total}}$. Resolve components at the start before applying any formula.

Pause, copy the highlighted sequence of four formulas into your book before moving on.

Fill the gap: For symmetrical projectile flight on level ground, the total flight time equals times the time to reach maximum height.

You’re here

Handle unequal heights and obstacles

Worked Example 3: Projectile Over a Wall

+5 XP

When launch and landing heights differ

We just saw the standard sequence for level-ground problems. That raises a question: what happens when the launch and landing heights are different, like over a cliff or past a wall? This card answers it → heights differ, so find time from $t = s_x/v_x$, then compute $s_y$ and add reference height.

PROBLEM 3 · UNEVEN GROUND

A ball is kicked at $18\ \text{m/s}$ at $35°$ above horizontal from the edge of a cliff that is 15 m above sea level. A wall stands 25 m horizontally from the cliff edge and is 8 m high (measured from sea level). Does the ball clear the wall? If so, by how much?

GIVEN: $v = 18\ \text{m/s}$, $\theta = 35°$, cliff $= 15\ \text{m}$, wall distance $= 25\ \text{m}$, wall height $= 8\ \text{m}$, $g = 9.8\ \text{m/s}^2$
Components: $v_x = 18\cos(35°) = 14.7\ \text{m/s}$, $v_y = 18\sin(35°) = 10.3\ \text{m/s}$

Launch and landing heights differ, the range equation does NOT apply. Use equations of motion.

When heights differ, the range equation does not apply. Find time from horizontal motion ($t = s_x/v_x$), compute vertical displacement ($s_y = v_y t - \tfrac{1}{2}g t^2$), add the reference height, then compare with any obstacle height.

Pause, write the highlighted uneven-ground procedure into your book.

Quick check: A stone is thrown from a cliff 30 m high. Why can’t you use $R = v^2\sin(2\theta)/g$ to find where it lands?

You’re here

Apply the protocol and check your answer

Key Formulas, Projectile Motion

$$s_x = v\cos\theta \cdot t \quad \text{(horizontal displacement)}$$

$$s_y = v\sin\theta \cdot t + \tfrac{1}{2}(-g)t^2 \quad \text{(vertical displacement)}$$

$$v_y = v\sin\theta - gt \quad \text{(vertical velocity at time } t\text{)}$$

$$h_{\text{max}} = \frac{(v\sin\theta)^2}{2g} \quad \text{(max height above launch)}$$

$$t_{\text{up}} = \frac{v\sin\theta}{g} \quad \text{(time to maximum height)}$$

$$R = \frac{v^2\sin(2\theta)}{g} \quad \text{equal launch/landing height; uniform }g\text{; negligible drag}$$

Real World: Olympic Shot Put

The shot put is launched from approximately 2.2 m height (the athlete’s shoulder) at roughly $13\ \text{m/s}$ and $38°$. The optimal angle is less than 45° because the launch is already elevated, a lower angle gives more horizontal speed while still allowing sufficient air time from the raised position.

Using the elevated launch model with the illustrative values $v = 13\ \text{m/s}$, $\theta = 38°$, $h_0 = 2.2\ \text{m}$:

$$s_y = -2.2 = 13\sin(38°)\cdot t - 4.9t^2 \Rightarrow t \approx 2.1\ \text{s}, \quad R \approx 24\ \text{m}$$

This gives an order-of-magnitude prediction for elite performance. A real throw also depends on release technique, aerodynamics and measurement uncertainty, so the ideal calculation is a model rather than a record prediction.

Activities · apply the protocol

Activity 1

ApplyBand 4

Finding the Launch Angle

An ideal projectile is launched at $28\ \text{m/s}$ and lands 68 m away at the launch height. Use the protocol to find both possible launch angles.

Hint: $\sin(2\theta)=Rg/v^2$. Sine has two solutions between $0°$ and $180°$.

Activity 2

ApplyBand 4–5

Projectile from a Balcony

A ball is thrown from a balcony 12 m above the ground at $15\ \text{m/s}$, $40°$ above horizontal. Find:

(a) the time to hit the ground
(b) the horizontal distance travelled
(c) the speed at impact
(d) the direction of the impact velocity, measured below the horizontal

Use $g = 9.8\ \text{m/s}^2$, an inertial ground frame, $+x$ forward and $+y$ upward. A complete velocity answer needs magnitude and direction.

Activity 3

AnalyseBand 5

Minimum Speed to Clear a Tree

A golfer wants to clear a tree that is 30 m away horizontally and 15 m high. The ball is struck from ground level. What is the minimum launch speed required at $45°$?

Hint: The ball must be at least 15 m high when $s_x = 30$ m. Set up simultaneous equations for horizontal and vertical motion.

Check the three activity answers

Activity 1: $\sin(2\theta)=68(9.8)/28^2=0.850$. Therefore $2\theta=58.2°$ or $121.8°$, giving $\theta=29.1°$ or $60.9°$. Both are valid in the stated ideal equal-height model.

Activity 2: $u_x=15\cos40°=11.5\ \text{m/s}$ and $u_y=15\sin40°=9.64\ \text{m/s}$. Solving $-12=9.64t-4.9t^2$ gives $t=2.83\ \text{s}$ (reject the negative root). Hence $s_x=32.5\ \text{m}$ and $v_y=9.64-9.8(2.83)=-18.1\ \text{m/s}$. The impact velocity is $21.4\ \text{m/s}$ at $57.6°$ below the horizontal.

Activity 3: At $\theta=45°$, $s_y=s_x-\dfrac{gs_x^2}{v^2}$. Setting $s_x=30\ \text{m}$ and $s_y=15\ \text{m}$ gives $v^2=588\ \text{m}^2\text{/s}^2$, so the minimum ideal launch speed is $24.2\ \text{m/s}$ (3 s.f.).

Motion Calculations

ApplyBand 3

Practise resolving and analysing motion

A projectile is launched at 18 m/s, 55° above the horizontal. Find the horizontal and vertical components of the initial velocity.

At a point in flight, the horizontal velocity is 12 m/s and the vertical velocity is 9 m/s. Find the resultant velocity (magnitude and direction).

A ball is thrown horizontally at 9 m/s from a cliff 39 m high. Calculate the time of flight and the horizontal distance travelled.

Copy into Books

Expand to see book notes

Title: Problem Solving: Projectiles

The Problem-Solving Protocol:

GIVEN List all knowns with units and signs; resolve velocity into components
FIND State the unknown
METHOD Write the equation and principle
ANSWER Substitute, calculate, check

Key Formulas: $s_x = u_x t$ · $s_y = u_y t + \frac{1}{2}a_y t^2$ · $v_y = u_y + a_y t$ · $h_{\text{max}} = u_y^2/(2g)$ · $t_{\text{up}} = u_y/g$.

Critical Check: $R=v^2\sin(2\theta)/g$ is a shortcut only for equal launch and landing heights, uniform $g$ and negligible drag. Otherwise solve the signed component equations.

Common Misconceptions

watch out

✗ “Use the range equation for all problems”

✓ Right: $R = v^2\sin(2\theta)/g$ only applies on level ground. Always check if launch and landing heights differ.

✗ “Time of flight depends on horizontal velocity”

✓ Right: Time is determined entirely by vertical motion. $v_x$ affects range but not flight time (ignoring air resistance).

✗ “At the top of the path, the velocity is zero”

✓ Right: At the peak, the vertical component of velocity is zero, but the horizontal component remains $v_x = v\cos\theta$. Gravity still acts.

Interactive Tool · Module 5 Projectile Solver

Revisit: Think First

Look back at the formulas you wrote at the start. Did you include all six key equations? Are your variable definitions and units correct? Add any you missed.

You’re here

Solve independently

Multiple Choice

+5 XP per correct · +25 XP all correct

A fresh set drawn from this lesson’s question bank, feedback shown immediately.

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer

+5 XP

ApplyBand 4–53 marks

Q1. A basketball player shoots at $8.5\ \text{m/s}$, $52°$ from horizontal from a height of $2.1\ \text{m}$. The basket is $3.05\ \text{m}$ high and $4.0\ \text{m}$ away horizontally. Determine whether the ball goes through the basket. Show all working using the problem-solving protocol. (3 marks)

AnalyseBand 54 marks

Q2. A stone is thrown from the top of a $60\ \text{m}$ cliff at $20\ \text{m/s}$, $40°$ above horizontal. Calculate:

(a) the time to reach the water
(b) the horizontal distance from the cliff base where it lands
(c) the impact velocity: magnitude and direction below the horizontal

Take $g = 9.8\ \text{m/s}^2$. (4 marks)

EvaluateBand 5–64 marks

Q3. Evaluate the problem-solving strategy of “always using the range equation first” when approaching projectile motion questions. Under what circumstances is this effective, and when does it lead to errors? (4 marks)

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 from the lesson bank.

Short Answer Model Answers

Q1 (3 marks):

GIVEN: $v = 8.5\ \text{m/s}$, $\theta = 52°$, $h_{\text{launch}} = 2.1\ \text{m}$, $h_{\text{basket}} = 3.05\ \text{m}$, $s_x = 4.0\ \text{m}$

FIND: Vertical position of the ball when $s_x = 4.0\ \text{m}$

METHOD: Find time from horizontal motion, then vertical displacement.

$t = s_x / (v\cos\theta) = 4.0 / (8.5 \times \cos 52°) = 4.0 / 5.24 = 0.764\ \text{s}$

$s_y = 8.5\sin 52° \times 0.764 + \frac{1}{2}(-9.8)(0.764)^2 = 6.70 \times 0.764 - 2.86 = 5.12 - 2.86 = 2.26\ \text{m}$

Total height above ground: $2.1 + 2.26 = 4.36\ \text{m}$. The basket is at $3.05\ \text{m}$. The ball is $4.36 - 3.05 = 1.31\ \text{m}$ above the basket. The ball overshoots the basket by approximately 1.3 m.

Q2 (4 marks):

(a) Take down as negative: $s_y = -60\ \text{m}$, $v_y = 20\sin 40° = 12.86\ \text{m/s}$

$-60 = 12.86t - 4.9t^2 \Rightarrow 4.9t^2 - 12.86t - 60 = 0$

$t = (12.86 + \sqrt{12.86^2 + 4(4.9)(60)}) / 9.8 = (12.86 + 36.6) / 9.8 = 5.05\ \text{s}$ (rejecting negative root)

(b) $v_x = 20\cos 40° = 15.32\ \text{m/s}$; $s_x = 15.32 \times 5.05 = 77.4\ \text{m}$

(c) $v_y = 12.86 - 9.8(5.05) = -36.6\ \text{m/s}$ and $v_x=15.32\ \text{m/s}$. Hence $|\vec v| = \sqrt{15.32^2 + 36.6^2}=39.7\ \text{m/s}$ and $\alpha=\tan^{-1}(36.6/15.32)=67.3°$. The impact velocity is $39.7\ \text{m/s}$ at $67.3°$ below the horizontal.

Q3 (4 marks):

Using the range equation first is effective only when: (i) launch and landing heights are the same; (ii) the question asks for range, launch speed, or launch angle; (iii) air resistance is negligible.

It leads to errors when: (i) launch and landing heights differ (most real-world problems); (ii) the question asks for maximum height or time of flight; (iii) there is an intermediate obstacle (e.g., does it clear a wall?).

Better strategy: Always resolve into components first, then use the appropriate equation of motion for the specific unknown. This works for all projectile problems, not just the special case of level ground.

You’re here

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.

Start the module quiz →

How did your thinking change?

At the start, a launch speed was identified as a vector magnitude rather than a horizontal or vertical component.

Reliable solutions begin with the frame: choose axes, attach signs, resolve the launch velocity, and state the model conditions. Only then choose an equation. This prevents a memorised shortcut from silently changing the physical situation.

I have completed this lesson

← Launch Angle Analysis Practical Investigation: Validating Projectile Motion →

Module overview · Physics · Checkpoint 1