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HSCScience Physics · Y12 · M5
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Year 12 Physics Module 5 ⏱ ~30 min 5 MC · 3 Short Answer Lesson 4 of 18

Phase 1 Consolidation

On 16 July 1969, NASA engineers at Mission Control solved 12 simultaneous projectile and orbital motion equations in real time during Apollo 11's trans-lunar injection burn. The 347-second burn accelerated the spacecraft from 7.79 km/s to 10.84 km/s — increasing velocity by 3.05 km/s with zero margin for error. Every calculation used the same kinematic tools you are about to consolidate.

Today's hook: On 16 July 1969, NASA Mission Control solved 12 simultaneous motion equations in real time during Apollo 11's trans-lunar injection burn — a 347-second burn that increased spacecraft velocity from 7.79 to 10.84 km/s. Before you consolidate today, write down the single hardest thing about projectile motion problems and what you'll focus on improving.
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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 Master Mastery tasks Exam Exam-style questions
Before you begin — reflect

What is the single hardest thing about projectile motion problems? What will you focus on improving?

Warm-up — which formula is used to find the range of a projectile launched and landing at the same height?

Learning Intentions
goals

Know — Recall all Phase 1 Formulae

  • Resolve velocity into components
  • Select the correct kinematic equation
  • Use range, max height, and flight time formulae

Understand — Identify and Correct Common Errors

  • Spot sign errors in vertical motion
  • Recognise when range formula does not apply
  • Fix component vs total velocity mistakes

Can Do — Solve Mixed Problems Under Exam Conditions

  • Complete 10 mixed practice questions
  • Answer 3 timed extended responses
  • Work accurately under time pressure
Key Terms — Projectile Motion
vocab
Velocity componentHorizontal ($v\cos\theta$) or vertical ($v\sin\theta$) part of a velocity vector.
Range (R)Horizontal displacement from launch to landing. Special formula only for level ground.
Maximum heightHighest point in the trajectory where $v_y = 0$.
Flight timeTotal time from launch to landing. Determined by vertical motion.
Complementary anglesTwo angles that add to 90°. They give the same range on level ground.
Cross-lesson links: L01–L03 built your projectile motion toolkit. L04 consolidates it — the mixed problem set here mirrors the HSC exam format: you will be given one of these problem types in Section II, worth 4–5 marks.
Misconceptions to fix before you review
✗ Wrong: You can use the range formula $R = v^2\sin(2\theta)/g$ whenever you have a launch angle and speed.
✓ Right: The range formula only applies when launch and landing heights are equal (level ground). From a cliff, use $s_y = u_y t + \tfrac{1}{2}at^2$ to find flight time, then $s_x = v_x t$.
✗ Wrong: At maximum height, the velocity of a projectile is zero.
✓ Right: At maximum height, only the vertical component is zero ($v_y = 0$). The horizontal component $v_x = v\cos\theta$ remains constant throughout the flight.

The range formula $R = v^2\sin(2\theta)/g$ applies when launching from a cliff.

At the maximum height of a projectile, both velocity and acceleration are zero.

IQ1: Projectile Motion Formula Sprint

1
Phase 1 Formula Sheet & Sprint Cards
+5 XP

Six formulae — when to use each, key variables, and common traps.

Phase 1 Summary

Phase 1 Summary — all six formulae at a glance.

Component equations

$v_x = v\cos\theta$   ·   $v_y = v\sin\theta$

Use: Resolving launch velocity into horizontal and vertical components

Displacement equations

$s_x = v_x t$   ·   $s_y = u_y t + \tfrac{1}{2}at^2$

Use: Horizontal displacement (constant velocity) and vertical displacement (constant acceleration)

Vertical velocity equations

$v_y = u_y + at$   ·   $v_y^2 = u_y^2 + 2as_y$

Use: Vertical velocity at time t, or linking velocity and vertical displacement without time

Range (level ground only)

$R = \dfrac{v^2\sin(2\theta)}{g}$

Use: Range on level ground only (launch height = landing height)

Maximum height

$h_{\max} = \dfrac{(v\sin\theta)^2}{2g}$

Use: Maximum height above launch point; derived from $v_y^2 = u_y^2 + 2as_y$ with $v_y = 0$

Flight time (level ground only)

$t_{\text{flight}} = \dfrac{2v\sin\theta}{g}$

Use: Total flight time on level ground only; time up = time down

Sprint Cards — click to reveal traps

$x = v_x \cdot t$
Click to reveal when to use, the trap, and connection
Variablesx = horizontal displacement, vx = horizontal velocity component = v cosθ (constant throughout flight), t = time of flight.
Use whenFinding horizontal range whenever the range formula does NOT apply (e.g., cliff launch, unequal heights). Always valid for horizontal motion.
TrapUse the horizontal component vx = v cosθ, not the total speed v. Time t comes from the vertical motion — flight time is determined by gravity, not horizontal speed.
$y = v_{y0}t + \tfrac{1}{2}gt^2$
Click to reveal when to use, the trap, and connection
Variablesy = vertical displacement, vy0 = initial vertical velocity component, g = acceleration due to gravity (−9.8 m/s² when up is positive), t = time.
Use whenFinding vertical displacement at a given time, or finding flight time from a cliff (set y = −h and solve the quadratic in t). Most versatile vertical equation.
TrapSign of g must match your convention. If up is positive, g = −9.8 m/s². Forgetting the negative is the most common vertical motion error.
$v_y = v_{y0} + gt$
Click to reveal when to use, the trap, and connection
Variablesvy = vertical velocity at time t, vy0 = initial vertical velocity, g = −9.8 m/s² (up positive).
Use whenYou want the vertical velocity component at any instant. Classic use: find vy at any point in flight, or find time to reach max height (set vy = 0).
TrapThis applies to vertical motion only. Horizontal velocity vx = v cosθ is constant throughout (a = 0 horizontally).
$v_y^2 = v_{y0}^2 + 2gy$
Click to reveal when to use, the trap, and connection
Variablesvy = vertical velocity, vy0 = initial vertical velocity, g = −9.8 m/s², y = vertical displacement. No time needed.
Use whenFinding maximum height (set vy = 0, solve for y) or finding vertical velocity after a given vertical displacement, without needing time.
TrapUse the vertical component vy0 = v sinθ, not the total launch speed v. Using total speed inflates the answer.
$R = \dfrac{v^2\sin(2\theta)}{g}$
Click to reveal when to use, the trap, and connection
VariablesR = horizontal range, v = launch speed, θ = launch angle above horizontal, g = 9.8 m/s².
Use whenLaunch and landing heights are equal (level ground). Fastest route to range when the problem gives speed and angle only. Maximum range occurs at θ = 45°.
TrapNEVER use from a cliff or raised platform (launch height ≠ landing height). In those cases use x = vx · t after finding t from the vertical equation. Complementary angles (e.g., 30° and 60°) give equal range.
$T = \dfrac{2v_{y0}}{g}$
Click to reveal when to use, the trap, and connection
VariablesT = total flight time, vy0 = initial vertical velocity = v sinθ, g = 9.8 m/s² (magnitude).
Use whenFinding total flight time on level ground only (launch and landing at the same height). Time up equals time down, so T = 2 × time to max height.
TrapOnly valid for level-ground launches. For cliff or unequal-height problems, solve y = vy0t + ½gt² for t instead.

The six projectile formulae are: $v_x = v\cos\theta$; $v_y = v\sin\theta$; $s_x = v_x t$; $s_y = u_y t + \tfrac{1}{2}at^2$; $v_y = u_y + at$; $v_y^2 = u_y^2 + 2as_y$. The range formula $R = v^2\sin(2\theta)/g$ and flight-time formula $t = 2v\sin\theta/g$ apply on level ground only — time always comes from the vertical motion.

Pause — copy all six highlighted formulae and the two level-ground-only conditions into your book before moving on.

A projectile is launched from the top of a cliff at 20 m/s, 30° above horizontal. Which formula gives the correct horizontal range?

Error Clinic

2
Six Common Projectile Errors
+5 XP

Each card shows a student's incorrect working. Find the error, explain the fix, then check.

We just saw all six projectile formulae and when each applies. That raises a question: which errors do students make most often when choosing or applying these formulae? This card answers it → six diagnostic cases, each revealing a different common trap.

1
Forgetting vx and vy are components
▾
Student working: A ball is thrown at 15 m/s at 30°. Find horizontal distance after 2 s.
Student writes: $s_x = vt = 15 \times 2 = 30$ m. Incorrect.
Fix: The student used the total velocity v = 15 m/s directly. They must first resolve into the horizontal component: $v_x = v\cos\theta = 15\cos 30° = 12.99$ m/s. Then $s_x = v_x t = 12.99 \times 2 = 26.0$ m. Always resolve velocity before using it in horizontal or vertical equations.
2
Using R = v²sin(2θ)/g when launch ≠ landing height
▾
Student working: A ball is kicked at 20 m/s, 40° from the edge of a 10 m cliff. Find where it lands.
Student uses: $R = \frac{20^2 \sin 80°}{9.8} = 40.2$ m. Incorrect.
Fix: The range formula $R = \frac{v^2\sin(2\theta)}{g}$ only applies when launch and landing heights are equal. From a cliff, the projectile falls further, so the range is larger. Use $s_y = u_y t + \frac{1}{2}at^2$ with $s_y = -10$ m to find flight time, then $s_x = v_x t$.
3
Sign errors — taking upward as positive but using a = +9.8
▾
Student working: A stone is thrown upward at 12 m/s. Find displacement after 1.5 s.
Student writes: $s_y = 12(1.5) + \frac{1}{2}(9.8)(1.5)^2 = 18 + 11.0 = 29.0$ m. Incorrect.
Fix: If upward is positive, then acceleration $a = -g = -9.8$ m s². The correct calculation: $s_y = 12(1.5) + \frac{1}{2}(-9.8)(1.5)^2 = 18 - 11.0 = +7.0$ m. The stone is 7.0 m above the launch point. Always define your sign convention and apply it consistently.
4
Using total velocity in vertical motion equations
▾
Student working: A projectile is launched at 25 m/s, 50°. Find maximum height.
Student writes: $v_y^2 = u_y^2 + 2as_y$ with $u_y = 25$ m/s, giving $0 = 25^2 + 2(-9.8)s_y$, so $h = 31.9$ m. Incorrect.
Fix: The student used total velocity v = 25 m/s instead of the vertical component $u_y = v\sin\theta = 25\sin 50° = 19.15$ m/s. Correct: $0 = (19.15)^2 + 2(-9.8)s_y$, so $h = \frac{367}{19.6} = 18.7$ m. Always use the vertical component in vertical motion equations.
5
Forgetting that vy = 0 at maximum height (not total v)
▾
Student working: A ball reaches maximum height. The student claims "the velocity is zero so it stops moving." Incorrect.
Fix: At maximum height, only the vertical component of velocity is zero ($v_y = 0$). The horizontal component $v_x = v\cos\theta$ remains constant throughout flight (ignoring air resistance). The projectile is still moving horizontally at its maximum height. The total velocity is equal to $v_x$, not zero.
6
Assuming time depends on horizontal velocity
▾
Student working: A ball is launched horizontally at 20 m/s from 45 m height. Find flight time.
Student writes: "Horizontal velocity is 20 m/s, so time depends on that." Then they get stuck. Incorrect reasoning.
Fix: Flight time is determined entirely by vertical motion, not horizontal velocity. For a horizontal launch, $u_y = 0$ and $s_y = \frac{1}{2}gt^2$. So $t = \sqrt{\frac{2s_y}{g}} = \sqrt{\frac{2 \times 45}{9.8}} = 3.03$ s. The horizontal velocity only affects the range ($s_x = v_x t$), not the time of flight.

The six common projectile errors: (1) using total v instead of $v_x$ or $v_y$; (2) applying the range formula when heights differ; (3) using $a = +9.8$ when up is positive; (4) using total v in vertical equations; (5) claiming total velocity is zero at max height (only $v_y = 0$); (6) thinking flight time depends on horizontal velocity.

Add the highlighted list of six errors to your notes before the check below.

Three of these statements about projectile motion are correct. Pick the odd one out (the incorrect statement).

Mixed Practice

3
Mixed Practice Questions
+5 XP

Ten questions from Band 3 to Band 6. Worked solutions in the Answers section of the Practice phase.

We just saw the six error patterns to avoid. That raises a question: can you now apply the correct formula to a range of mixed problems without falling into those traps? This card answers it → ten graded questions from Band 3 to Band 6.

Apply Band 3 2 marks

Q1. A ball is thrown at 12 m/s, 40° above horizontal. Find the horizontal and vertical components of the initial velocity.

Apply Band 3 2 marks

Q2. A projectile is launched horizontally at 15 m/s from a height of 40 m. Find: (a) the time to fall to the ground, and (b) the horizontal range.

Understand Band 3 2 marks

Q3. State two assumptions made in deriving the range formula $R = \frac{v^2\sin(2\theta)}{g}$.

Apply Band 4/5 3 marks

Q4. A ball is kicked at 22 m/s, 35° above horizontal on level ground. Find: (a) the maximum height reached, (b) the total flight time, and (c) the horizontal range.

Apply Band 4/5 3 marks

Q5. A stone is projected from the top of a 25 m cliff at 18 m/s, 30° above horizontal. Find the horizontal distance from the base of the cliff where the stone lands.

Analyse Band 4/5 3 marks

Q6. Show that launch angles of 20° and 70° give the same range on level ground for the same launch speed.

Analyse Band 4/5 4 marks

Q7. A projectile must clear a 5 m wall located 15 m away. It is launched from ground level at 25°. Find the minimum launch speed required.

Analyse Band 6 4 marks

Q8. A ball is thrown from 1.5 m above the ground and must land 20 m away in a basket 3.0 m high. Find the launch speed and launch angle required.

Analyse Band 6 4 marks

Q9. Two projectiles A and B are launched at the same speed. A at 30°, B at 60°. Compare their maximum heights, flight times, and ranges. Explain your answers using physics principles.

Analyse Band 6 5 marks

Q10. A firework is launched from ground level at $v$ m/s, $\theta$°. At its maximum height it explodes, sending sparks horizontally at 5 m/s. One spark lands 45 m from the launch point. Given the total time from launch to spark landing is 4.5 s, find $v$ and $\theta$.

For every projectile problem: (1) resolve into components; (2) check launch vs landing height; (3) for cliff problems, set $s_y = -h$ and solve for $t$, then use $s_x = v_x t$; (4) complementary angles $\theta$ and $(90° - \theta)$ give equal range because $\sin(2\theta) = \sin(180° - 2\theta)$.

Pause — write the highlighted four-step process and the cliff problem method into your book.

Quick check — for Q2 above (horizontal launch from 40 m), which equation correctly gives the flight time?

Timed Exam Block

4
Extended Response Practice
+5 XP

Three exam-style questions. Each: 4 marks, 8 minutes recommended. Solutions in the Answers section of the Practice phase.

We just saw ten mixed practice questions building fluency with all six formulae. That raises a question: can you now sustain accuracy under timed, exam-like pressure? This card answers it → three extended-response questions, each 4 marks and 8 minutes.

08:00
Question 11 (4 marks, ~8 min)

A long jumper takes off at 9.2 m/s, 22° above horizontal.

  1. Calculate the horizontal and vertical components of the take-off velocity. (1 mark)
  2. Calculate the time of flight. (1 mark)
  3. Calculate the horizontal distance jumped. (1 mark)
  4. Calculate the maximum height above the take-off point. (1 mark)
08:00
Question 12 (4 marks, ~8 min)

A ball is thrown from the top of a 30 m building at 15 m/s, 50° above horizontal.

  1. Calculate the time to reach maximum height. (1 mark)
  2. Calculate the total flight time. (1 mark)
  3. Calculate the horizontal distance travelled from the base of the building. (1 mark)
  4. Calculate the velocity at impact (magnitude and direction). (1 mark)
08:00
Question 13 (4 marks, ~8 min)

A projectile is launched at 30 m/s. On level ground it travels 72 m.

  1. Find the two possible launch angles. (1 mark)
  2. Find the flight time for each angle. (1 mark)
  3. Explain why there are two possible angles that give the same range. (2 marks)

Exam technique: for every extended-response projectile question, state GIVEN/FIND/METHOD/ANSWER, resolve components in step 1, check level-ground condition, then work vertically and horizontally in separate labelled steps.

Add the highlighted exam technique reminder to your notes before the activities.

Activity 1 — Projectile Motion Consolidation Drills
ApplyBand 3

Practise the key projectile motion concepts from this phase.

  1. State the range formula $R = v^2\sin(2\theta)/g$ and list two conditions that must be satisfied for it to be valid.
  2. A ball is launched at 25 m/s at 40° above horizontal on level ground. Use the range formula to calculate the horizontal range. Show all working.
  3. Explain why the time of flight $T = 2v\sin\theta/g$ depends only on the vertical component of velocity, and give a real-world example that illustrates this principle.
Activity 2 — Concept Check
UnderstandBand 5

Explain the reasoning behind a key projectile motion principle.

A student claims that a ball thrown at 30° and a ball thrown at 60° (same speed, level ground) land at the same point but travel completely different paths. Explain why this is true using the range formula $R = v^2\sin(2\theta)/g$ and describe the key differences between the two trajectories (maximum height, flight time, horizontal speed).

Fill the gap: launch angles of 30° and 60° give the same range because $\sin(2 \times 30°) = \sin(\_\_\_°)$, which equals $\sin(2 \times 60°)$. The missing value is _____.

Wrap-up — Misconceptions & Summary

Misconceptions — final check

✗ "I can use the range formula for any projectile problem."
✓ The range formula only works on level ground (launch height = landing height). For cliff or uneven ground, always use the vertical displacement equation to find time first.
✗ "The larger the launch angle, the further the projectile goes."
✓ Range depends on $\sin(2\theta)$, which is maximum at 45°. For $\theta > 45°$, range actually decreases. Complementary angles (e.g., 30° and 60°) give identical ranges.

Copy into your books

The Six Formulae

  • $v_x = v\cos\theta$ · $v_y = v\sin\theta$
  • $s_x = v_x t$ · $s_y = u_y t + \tfrac{1}{2}at^2$
  • $v_y = u_y + at$ · $v_y^2 = u_y^2 + 2as_y$

Level-Ground Only

  • $R = \dfrac{v^2\sin(2\theta)}{g}$
  • $h_{\max} = \dfrac{(v\sin\theta)^2}{2g}$
  • $t_{\text{flight}} = \dfrac{2v\sin\theta}{g}$

Six Common Errors

  • Using total v instead of $v_x$ or $v_y$
  • Range formula for cliff problems
  • Wrong sign for $g$ (up positive → $a = -9.8$)

Key Principles

  • Horizontal: $a=0$, $v_x$ constant
  • Vertical: $a = -9.8$ m/s² (up positive)
  • Time set by vertical motion only

Match each scenario to the correct approach.

Quick recall — Phase 1 Consolidation
+5 XP

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.

Short Answer — 10 marks
+5 XP

AnalyseBand 5(3 marks) 1. A student claims that increasing the launch angle of a projectile always increases its range. Assess this claim with reference to the range formula and specific angles.

1 mark: states range depends on $\sin(2\theta)$ max at 45° · 1 mark: explains range decreases beyond 45° · 1 mark: specific angle example (e.g., 30° and 60° give same range)

ApplyBand 4(3 marks) 2. A ball is thrown at 16 m/s at an angle of 40° above horizontal. Calculate the horizontal and vertical components of velocity, the time to reach maximum height, and the maximum height above the launch point.

1 mark: correct $v_x$ and $v_y$ · 1 mark: time to max height using $v_y = 0$ · 1 mark: max height using energy or kinematic equation

EvaluateBand 6(4 marks) 3. Evaluate the statement: "The time of flight of a projectile depends only on its vertical component of velocity and is independent of the horizontal component." Use equations and a numerical example to justify your answer.

1 mark: statement is correct · 1 mark: flight time formula shown · 1 mark: numerical example · 1 mark: explains horizontal independence

Show all answers

Mixed Practice Answers (Q1–Q10)

Q1 (2 marks): $v_x = v\cos\theta = 12\cos 40° = \mathbf{9.19 \text{ m/s}}$ (1 mark). $v_y = v\sin\theta = 12\sin 40° = \mathbf{7.71 \text{ m/s}}$ (1 mark).

Q2 (2 marks): (a) Vertical: $s_y = \frac{1}{2}gt^2$ so $t = \sqrt{\frac{2s_y}{g}} = \sqrt{\frac{2 \times 40}{9.8}} = \mathbf{2.86 \text{ s}}$ (1 mark). (b) Horizontal: $s_x = v_x t = 15 \times 2.86 = \mathbf{42.9 \text{ m}}$ (1 mark).

Q3 (2 marks): Any two of: (1) No air resistance; (2) Constant $g = 9.8$ m s² downward; (3) Launch and landing heights are equal (level ground); (4) Flat Earth (g is uniform). (1 mark each).

Q4 (3 marks): $u_y = 22\sin 35° = 12.62$ m/s; $v_x = 22\cos 35° = 18.02$ m/s. (a) $h_{\max} = \frac{u_y^2}{2g} = \frac{12.62^2}{2 \times 9.8} = \mathbf{8.13 \text{ m}}$ (1 mark). (b) $t_{\text{flight}} = \frac{2u_y}{g} = \frac{2 \times 12.62}{9.8} = \mathbf{2.58 \text{ s}}$ (1 mark). (c) $R = 18.02 \times 2.58 = \mathbf{46.5 \text{ m}}$ (1 mark).

Q5 (3 marks): $u_y = 18\sin 30° = 9.0$ m/s; $v_x = 18\cos 30° = 15.59$ m/s. Vertical: $-25 = 9.0t - 4.9t^2$ → $4.9t^2 - 9.0t - 25 = 0$. $t = \frac{9.0 + \sqrt{81 + 490}}{9.8} = \mathbf{3.35 \text{ s}}$ (2 marks). Horizontal: $s_x = 15.59 \times 3.35 = \mathbf{52.2 \text{ m}}$ (1 mark).

Q6 (3 marks): For θ = 20°: $R_1 = \frac{v^2\sin(40°)}{g}$ (1 mark). For θ = 70°: $R_2 = \frac{v^2\sin(140°)}{g}$ (1 mark). Since $\sin(140°) = \sin(180° - 40°) = \sin(40°)$, we have $R_1 = R_2$ (1 mark).

Q7 (4 marks): At wall: $t = \frac{15}{v\cos 25°}$ (1 mark). Substituting into $s_y = v\sin\theta \cdot t - \frac{1}{2}gt^2$: $5 = 15\tan 25° - \frac{4.9 \times 225}{v^2\cos^2 25°}$. $\frac{1102.5}{0.821v^2} = 1.99$ (1 mark). $v = \mathbf{26.0 \text{ m/s}}$ (2 marks).

Q8 (4 marks): $\Delta s_x = 20$ m, $\Delta s_y = +1.5$ m. $\tan\theta = \frac{1.5 + 4.9t^2}{20}$. Try $t = 1.5$ s: $\theta \approx 32.0°$, $v = 15.7$ m/s. Check: $s_y = 15.7\sin 32° \times 1.5 - 4.9(1.5)^2 = 1.5$ m ✓. $\mathbf{\theta \approx 32°, v \approx 15.7 \text{ m/s}}$ (4 marks for method).

Q9 (4 marks): Same speed $v$, A at 30°, B at 60°. Height: $h \propto \sin^2\theta$. B reaches 3× the height of A (1 mark). Flight time: $t \propto \sin\theta$. B stays airborne 1.73× longer (1 mark). Range: $R \propto \sin(2\theta)$. Same range — complementary angles (1 mark). Explanation: Smaller angle → less vertical velocity (shorter time, lower height) but more horizontal velocity. These balance exactly for range (1 mark).

Q10 (5 marks): $t_1 = t_2 = 2.25$ s (1 mark each for method + value). $v\sin\theta = 9.8 \times 2.25 = 22.05$ m/s. $v\cos\theta = 15$ m/s (1 mark). $v = \sqrt{22.05^2 + 15^2} = \mathbf{26.7 \text{ m/s}}$. $\theta = \tan^{-1}(22.05/15) = \mathbf{55.8°}$ (2 marks).

Timed Exam Answers (Q11–Q13)

Q11 (4 marks): (a) $v_x = 9.2\cos 22° = \mathbf{8.53 \text{ m/s}}$; $u_y = 9.2\sin 22° = \mathbf{3.45 \text{ m/s}}$ (1 mark). (b) $t = \frac{2u_y}{g} = \frac{2 \times 3.45}{9.8} = \mathbf{0.704 \text{ s}}$ (1 mark). (c) $R = 8.53 \times 0.704 = \mathbf{6.01 \text{ m}}$ (1 mark). (d) $h_{\max} = \frac{u_y^2}{2g} = \frac{3.45^2}{19.6} = \mathbf{0.607 \text{ m}}$ (1 mark).

Q12 (4 marks): $v_x = 15\cos 50° = 9.64$ m/s; $u_y = 15\sin 50° = 11.49$ m/s. (a) $t_{up} = \frac{u_y}{g} = \frac{11.49}{9.8} = \mathbf{1.17 \text{ s}}$ (1 mark). (b) $-30 = 11.49t - 4.9t^2$ → $t = \mathbf{3.92 \text{ s}}$ (1 mark). (c) $s_x = 9.64 \times 3.92 = \mathbf{37.8 \text{ m}}$ (1 mark). (d) $v_y = 11.49 - 9.8(3.92) = -26.9$ m/s. $v = \sqrt{9.64^2 + 26.9^2} = \mathbf{28.6 \text{ m/s}}$ at 70.3° below horizontal (1 mark).

Q13 (4 marks): (a) $\sin(2\theta) = \frac{72 \times 9.8}{900} = 0.784$. $\theta = \mathbf{25.8°}$ or $\mathbf{64.2°}$ (1 mark). (b) $t_{25.8°} = \mathbf{2.66 \text{ s}}$; $t_{64.2°} = \mathbf{5.52 \text{ s}}$ (1 mark). (c) Two angles because $\sin(2\theta)$ has the same value for $2\theta$ and $(180° - 2\theta)$. Lower angle: less vertical velocity (shorter time) but more horizontal velocity. Higher angle: more vertical velocity (longer time) but less horizontal speed. These combinations produce the same range (2 marks).

Multiple Choice — Key

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

SA1 (3 marks): The claim is incorrect. From $R = \frac{v^2\sin(2\theta)}{g}$, range depends on $\sin(2\theta)$, which is maximum at $\theta = 45°$ (1 mark). Increasing the angle from 30° to 45° does increase range. But increasing beyond 45° decreases range because $\sin(2\theta)$ decreases (1 mark). For example, $\sin(60°) = 0.866$ at 30° and $\sin(120°) = 0.866$ at 60° — same range as 30°. So range peaks at 45° then decreases (1 mark).

SA2 (3 marks): $v_x = 16\cos 40° = \mathbf{12.3 \text{ m/s}}$; $u_y = 16\sin 40° = \mathbf{10.3 \text{ m/s}}$ (1 mark). $t_{up} = \frac{u_y}{g} = \frac{10.3}{9.8} = \mathbf{1.05 \text{ s}}$ (1 mark). $h_{\max} = \frac{u_y^2}{2g} = \frac{10.3^2}{19.6} = \mathbf{5.41 \text{ m}}$ (1 mark).

SA3 (4 marks): The statement is correct (1 mark). Flight time from $T = \frac{2v\sin\theta}{g} = \frac{2u_y}{g}$ depends only on $u_y$ (1 mark). Numerical example: launch at 20 m/s at 30° vs 60°. $t_{30} = \frac{2 \times 20 \times 0.5}{9.8} = 2.04$ s; $t_{60} = \frac{2 \times 20 \times 0.866}{9.8} = 3.54$ s. Horizontal components changed but flight time changed because vertical component changed (1 mark). Even if $v_x$ doubled while $u_y$ stayed the same, flight time would not change (1 mark).

Boss Battle — Module Quiz
boss

Five timed questions on Phase 1 projectile motion. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

⚔ Enter the arena
Arcade Practice · Doodle Jump

Keep your momentum going — a quick game between study sessions keeps the mind sharp.

Revisit — How did your thinking change?

At the start you were asked about NASA Mission Control's 16 July 1969 Apollo 11 trans-lunar injection — 12 simultaneous equations solved in real time during a 347-second burn that changed velocity from 7.79 to 10.84 km/s.

Every one of those equations used the kinematic tools in L01–L03: component resolution, time-of-flight calculation, and applying the correct equation of motion to the correct direction. The same skills you are consolidating today. After working through the practice questions, reflect:

  • Which question type do you find easiest? Which is hardest?
  • Which error from the Error Clinic have you made before?
  • What is your plan to avoid that error in the exam?
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Module overview · Physics · Checkpoint 1 · Module quiz

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