Physics • Year 12 • Module 5 • Lesson 2
Launch Angle Analysis
Lock in the range equation, the 45° maximum, and the complementary angle property before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: range, complementary angles, double-angle identity, optimal angle, time of flight, level ground, maximum range, launch speed. 8 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The total horizontal displacement of a projectile from launch to landing. | |
| 1.2 | Two launch angles that sum to 90°; they always produce the same range on level ground. | |
| 1.3 | The trigonometric identity sin(2θ) = 2 sinθ cosθ used to simplify the range formula. | |
| 1.4 | The launch angle that produces the greatest range for a given launch speed and height condition. | |
| 1.5 | The total time a projectile spends in the air from launch until landing. | |
| 1.6 | The condition in which launch height equals landing height, so the net vertical displacement is zero. | |
| 1.7 | The greatest possible range, equal to v²/g on level ground; achieved at 45°. | |
| 1.8 | The initial speed of the projectile at the moment of launch; denoted v. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 On level ground, a projectile launched at 45° always achieves the greatest possible range for that launch speed. T / F
2.2 Launch angles of 20° and 80° produce the same range on level ground because they sum to 100°. T / F
2.3 The range equation R = v² sin(2θ)/g applies when there is no air resistance and launch and landing are at the same height. T / F
2.4 When a projectile is launched from a cliff, the optimal angle for maximum range is greater than 45°. T / F
2.5 The maximum range on level ground depends only on launch speed and gravitational acceleration; it does not depend on mass. T / F
2.6 The time of flight on level ground is given by t = v sinθ / g. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
45° · complementary · double-angle · g · level · less than · range · sin(2θ)
The horizontal displacement from launch to landing is called the ___________. For a projectile launched at speed v on ___________ ground, the range equation is R = v² ___________ / ___________. This formula uses the ___________ identity to replace 2 sinθ cosθ with a single trigonometric term. The maximum range occurs at a launch angle of ___________ because that angle maximises sin(2θ). Two angles that sum to 90° are called ___________ angles; they always produce the same range. When a projectile is launched from a height above the landing point, the optimal angle is ___________ 45°.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 State the range equation for a projectile on level ground and identify every variable.
4.2 Explain why 45° gives the maximum range on level ground, using the double-angle identity in your answer.
4.3 State the formula for time of flight on level ground and explain what happens to the time of flight when the launch angle increases from 30° to 60°.
4.4 Explain why launching from a cliff shifts the optimal angle to less than 45°. Refer to horizontal velocity and flight time in your answer.
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “is maximised at”, “is derived from”, “always give equal”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: range equation · 45° · complementary angles · maximum range · time of flight · launch speed.
Q1 — Term–definition match
1.1 range • 1.2 complementary angles • 1.3 double-angle identity • 1.4 optimal angle • 1.5 time of flight • 1.6 level ground • 1.7 maximum range • 1.8 launch speed.
Q2 — True / false with correction
2.1 True. sin(2 × 45°) = sin(90°) = 1, the maximum value of sin(2θ), so R = v²/g is the greatest possible range.
2.2 False. Complementary angles must sum to 90°, not 100°. The correct pair that produces the same range as 20° is 70° (since 20° + 70° = 90°). Angles that sum to 100° do not give equal ranges.
2.3 True.
2.4 False. When launched from a cliff (above the landing point), the projectile has extra fall time. A shallower trajectory (less than 45°) makes better use of this extended flight. The optimal angle is less than 45°, not greater.
2.5 True. Rmax = v²/g depends only on v and g. In vacuum, mass cancels from all projectile equations.
2.6 False. The time of flight on level ground is t = 2v sinθ / g — the factor of 2 is essential because the projectile travels up and then back down to the same height.
Q3 — Cloze paragraph
In order: range • level • sin(2θ) • g • double-angle • 45° • complementary • less than.
Q4.1 — Range equation
R = v² sin(2θ) / g, where R is the range (m), v is the launch speed (m/s), θ is the launch angle above the horizontal (degrees), and g is gravitational acceleration (9.8 m/s²). Valid only on level ground with no air resistance.
Q4.2 — Why 45° gives maximum range
Using the double-angle identity, R = v² sin(2θ) / g. The factor sin(2θ) reaches its maximum value of 1 when 2θ = 90°, i.e. θ = 45°. At that angle, all the range equation’s trigonometric factor is unity, so R = v²/g, which is the largest possible value for any θ.
Q4.3 — Time of flight and changing angle
Time of flight on level ground: t = 2v sinθ / g. As θ increases from 30° to 60°, sinθ increases (sin 30° = 0.500; sin 60° = 0.866), so the time of flight increases. A steeper launch sends the projectile higher, extending the time before it returns to ground level.
Q4.4 — Optimal angle from a cliff
Launching from a cliff gives the projectile extra time to fall to the (lower) landing point. A shallower angle (below 45°) sacrifices some height but retains a larger horizontal velocity component; the greater vx = v cosθ then acts over the extended flight time to produce a longer range than a 45° launch would achieve.
Q5 — Sample concept map
Correct maps should include arrows such as:
- range equation — uses → launch speed
- range equation — uses → time of flight
- 45° — is the optimal angle for → maximum range
- maximum range — equals v²/g using → launch speed
- complementary angles — always produce equal → range equation output
- time of flight — increases with θ toward → 45°
Award 1 mark per valid labelled arrow with a meaningful linking phrase (minimum 6).