Launch Angle Analysis
In 1918, the German Krupp Paris-Geschütz — the largest artillery piece ever built — fired 367 shells at Paris from 130 km away. Each 94 kg shell took approximately 170 s in flight. Yet the gun fired at 55°, not 45°, because Earth's rotation and air resistance shifted the optimal angle. Understanding the range equation reveals exactly why 45° is only optimal under ideal conditions.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Estimate the range of a soccer penalty kick struck at 25 m/s, 20° above horizontal. Show your reasoning.
Write your estimate before working through the lesson — you will revisit it at the end.
Warm-up — for a projectile on level ground, what determines the time of flight?
Know — Range Equation
- Derive and use the range equation for projectile motion on level ground
Understand — Maximum Range
- Explain why 45° gives maximum range on level ground using the sine double-angle identity
Can Do — Height Effects
- Analyse how different launch and landing heights affect the optimal launch angle
Core Content
Deriving a compact formula for horizontal range on level ground
Fire a cannonball from level ground at 100 m/s. At 30° it lands closer than at 45°; at 60° it also lands closer than at 45° but travels the same distance as the 30° shot. There is a single angle that maximises horizontal range when launch and landing height are equal — and deriving why requires combining the horizontal and vertical motion equations.
Launch Angle Analysis — how angle affects range on level ground.
Derivation
The horizontal velocity is constant (no air resistance):
$v_x = v\cos\theta$
The time of flight is found from vertical motion. On level ground, the projectile returns to its initial vertical position, so $s_y = 0$:
$s_y = v_y t + \tfrac{1}{2}a_y t^2$
$0 = v\sin\theta \cdot t_{\text{flight}} - \tfrac{1}{2}g \cdot t_{\text{flight}}^2$
$t_{\text{flight}} = \dfrac{2v\sin\theta}{g}$
The range is horizontal velocity multiplied by time of flight:
$R = v_x \times t_{\text{flight}} = v\cos\theta \times \dfrac{2v\sin\theta}{g} = \dfrac{v^2(2\sin\theta\cos\theta)}{g}$
Using the double-angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$:
$R = \dfrac{v^2\sin(2\theta)}{g}$
Maximum Range
The maximum value of $\sin(2\theta)$ is 1, which occurs when $2\theta = 90°$, giving $\theta = 45°$.
At $\theta = 45°$:
$R_{\text{max}} = \dfrac{v^2}{g}$
This is a remarkable result: the maximum range depends only on launch speed and gravitational acceleration, not on any other parameter.
Given: A ball is kicked on level ground at 15 m/s. Find the range at launch angles of 30°, 45°, and 60°. Take $g = 9.8\ \text{m/s}^2$.
- Find. Range $R$ at three different launch angles.
- Method. Use $R = v^2\sin(2\theta)/g$ with $v = 15\ \text{m/s}$ for each angle.
- At 30°. $R = \dfrac{(15\ \text{m/s})^2 \times \sin(60°)}{9.8\ \text{m/s}^2} = \dfrac{225 \times 0.866}{9.8}\ \text{m} = 19.9\ \text{m}$
- At 45°. $R = \dfrac{(15\ \text{m/s})^2 \times \sin(90°)}{9.8\ \text{m/s}^2} = \dfrac{225 \times 1}{9.8}\ \text{m} = 23.0\ \text{m}$
- At 60°. $R = \dfrac{(15\ \text{m/s})^2 \times \sin(120°)}{9.8\ \text{m/s}^2} = \dfrac{225 \times 0.866}{9.8}\ \text{m} = 19.9\ \text{m}$
Answer: At 30° the range is 19.9 m; at 45° it is 23.0 m (maximum); at 60° it is 19.9 m. The 30° and 60° launches give identical ranges — a preview of complementary angles.
On level ground the range is $R = v^2\sin(2\theta)/g$; it is maximised at $\theta = 45°$, giving $R_{\text{max}} = v^2/g$. The time of flight is $t_{\text{flight}} = 2v\sin\theta/g$ and is derived by setting $s_y = 0$.
Pause — copy the highlighted range equation and the maximum-range result into your book before moving on.
A ball is kicked on level ground at 20 m/s. What is the maximum possible range? ($g = 9.8\ \text{m/s}^2$)
Two different launch angles that produce exactly the same range
We just saw that the range equation contains $\sin(2\theta)$, which peaks at 45°. That raises a question: are there other angles that give the same range as each other? This card answers it → complementary angles (summing to 90°) produce identical ranges.
A powerful symmetry in the range equation reveals that two different launch angles can produce exactly the same range.
Recall the identity: $\sin(180° - \phi) = \sin(\phi)$. If we set $\phi = 2\theta$, then:
$\sin(2\theta) = \sin(180° - 2\theta) = \sin\bigl(2(90° - \theta)\bigr)$
This means launching at angle $\theta$ gives the same range as launching at angle $(90° - \theta)$. These angles are called complementary angles (they sum to 90°).
30° and 60° give the same range because $\sin(60°) = \sin(120°)$. The steeper trajectory (60°) goes higher but flies for less time; the shallower trajectory (30°) stays lower but flies for longer. The horizontal distance covered is identical.
Given: A projectile is launched on level ground at 25 m/s. Show that launch angles of 25° and 65° give the same range. Take $g = 9.8\ \text{m/s}^2$.
- Find. Show that $R_{25°} = R_{65°}$ using the range equation.
- Method. Use $R = v^2\sin(2\theta)/g$. For 25°, compute $\sin(50°)$. For 65°, compute $\sin(130°)$. Show these sines are equal.
- At 25°. $R = \dfrac{(25\ \text{m/s})^2 \times \sin(50°)}{9.8\ \text{m/s}^2} = \dfrac{625 \times 0.766}{9.8}\ \text{m} = 48.9\ \text{m}$
- At 65°. $R = \dfrac{(25\ \text{m/s})^2 \times \sin(130°)}{9.8\ \text{m/s}^2} = \dfrac{625 \times 0.766}{9.8}\ \text{m} = 48.9\ \text{m}$
- Verification. Since $\sin(50°) = \sin(130°) = 0.766$, and $25° + 65° = 90°$, the two angles are complementary and produce identical ranges.
Answer: Both 25° and 65° give a range of 48.9 m, confirming that complementary launch angles produce the same range on level ground.
Complementary angles $\theta$ and $(90° - \theta)$ produce identical ranges on level ground because $\sin(2\theta) = \sin(180° - 2\theta)$; for example, 30° and 60° always give the same range at any launch speed.
Add the highlighted complementary-angle rule to your notes before the check below.
30° and 60° launch angles produce the same range on level ground at any given speed.
Complementary angles are any two angles that give different ranges.
A 25° and 65° launch at the same speed produce the same range because $\sin(50°) = \sin(130°)$.
How height difference shifts the optimal launch angle away from 45°
We just saw that 45° maximises range on level ground and complementary angles give equal ranges. That raises a question: does 45° stay optimal when launch and landing heights differ? This card answers it → height difference shifts the optimal angle below 45° (from a cliff) or above 45° (uphill).
So far we have assumed launch and landing at the same height. In many real situations this is not true — a projectile may be launched from a cliff, a ramp, or a building, or may land on a slope.
Time of Flight from Height
When a projectile is launched from height $h$ above the landing point, the vertical displacement is $s_y = -h$ (downward). Using the quadratic formula on $s_y = v_y t + \tfrac{1}{2}a_y t^2$:
$-h = v\sin\theta \cdot t - \tfrac{1}{2}g \cdot t^2$
$\tfrac{1}{2}gt^2 - v\sin\theta \cdot t - h = 0$
Solving this quadratic for the positive root:
$t_{\text{flight}} = \dfrac{v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh}}{g}$
The extra term $+2gh$ under the square root increases the flight time compared to level ground. The range is then $R = v\cos\theta \times t_{\text{flight}}$.
Why the Optimal Angle Changes
The projectile has extra time to fall. A shallower angle (less than 45°) lets the horizontal velocity component carry it further during this extended flight. Optimal angle is less than 45°.
The projectile lands sooner. A steeper angle (greater than 45°) gives more vertical velocity to reach the higher landing point. Optimal angle is greater than 45°.
Given: A projectile is launched from a 30 m cliff at 20 m/s. Calculate the range at 30°, 45°, and 60°. Take $g = 9.8\ \text{m/s}^2$.
- Find. Range at three angles from a height of 30 m above landing.
- Method. Use $t_{\text{flight}} = (v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh})/g$, then $R = v\cos\theta \times t_{\text{flight}}$.
- At 30°.
$v_y = 20\sin(30°) = 10.0\ \text{m/s}$
$t = \dfrac{10.0 + \sqrt{100 + 2 \times 9.8 \times 30}}{9.8}\ \text{s} = \dfrac{10.0 + \sqrt{688}}{9.8}\ \text{s} = \dfrac{10.0 + 26.2}{9.8}\ \text{s} = 3.70\ \text{s}$
$R = 20\cos(30°) \times 3.70 = 17.3 \times 3.70\ \text{m} = 64.0\ \text{m}$ - At 45°.
$v_y = 20\sin(45°) = 14.1\ \text{m/s}$
$t = \dfrac{14.1 + \sqrt{199 + 588}}{9.8}\ \text{s} = \dfrac{14.1 + 28.0}{9.8}\ \text{s} = 4.30\ \text{s}$
$R = 20\cos(45°) \times 4.30 = 14.1 \times 4.30\ \text{m} = 60.6\ \text{m}$ - At 60°.
$v_y = 20\sin(60°) = 17.3\ \text{m/s}$
$t = \dfrac{17.3 + \sqrt{299 + 588}}{9.8}\ \text{s} = \dfrac{17.3 + 29.7}{9.8}\ \text{s} = 4.80\ \text{s}$
$R = 20\cos(60°) \times 4.80 = 10.0 \times 4.80\ \text{m} = 48.0\ \text{m}$
Answer: At 30°: 64.0 m; at 45°: 60.6 m; at 60°: 48.0 m. From a cliff, the optimal angle is less than 45° — in this case, 30° gives the greatest range.
When launched from height $h$, flight time is $t = (v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh})/g$; launching above landing makes the optimal angle less than 45°, and launching below landing (uphill) makes it greater than 45°.
Pause — write the highlighted cliff-launch formula and the two height-effect rules into your book.
Three of these statements about a projectile launched from a cliff are correct. Pick the odd one out.
Range (level ground): $R = \dfrac{v^2\sin(2\theta)}{g}$
Time of flight (level ground): $t_{\text{flight}} = \dfrac{2v\sin\theta}{g}$
Time of flight from height $h$: $t_{\text{flight}} = \dfrac{v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh}}{g}$
Maximum range (level ground): $R_{\text{max}} = \dfrac{v^2}{g}$ at $\theta = 45°$
Fill the gap. For a projectile launched from a cliff of height $h$, the flight time formula has an extra term under the square root: $t = (v\sin\theta + \sqrt{v^2\sin^2\theta + \_\_\_gh})/g$. The missing coefficient is _____.
A penalty kick is taken 11 m from the goal line, struck at approximately 25 m/s. At 15° launch angle: the range is approximately 32 m — easily reaching the goal with room to spare. At 45°: the range would be approximately 64 m, far overshooting the goal.
Players deliberately use low launch angles for accuracy and control, not maximum range. The crossbar is 2.44 m high, so the trajectory must stay below this while still beating the goalkeeper. Optimal angle in sport is almost never 45° — accuracy, speed constraints, and launch height all play a role.
Why do soccer players use a low launch angle (around 15°) for penalty kicks rather than 45°?
Complete the table for a projectile launched at $v = 20\ \text{m/s}$ on level ground. Take $g = 9.8\ \text{m/s}^2$.
| Launch Angle $\theta$ | $\sin(2\theta)$ | Range $R$ (m) | $t_{\text{flight}}$ (s) |
|---|---|---|---|
| 15° | 0.500 | ________ | ________ |
| 30° | 0.866 | ________ | ________ |
| 45° | 1.000 | ________ | ________ |
| 60° | 0.866 | ________ | ________ |
| 75° | 0.500 | ________ | ________ |
Questions:
- Which angle gives the maximum range? By how much does it exceed the range at 15°?
- Which pairs of angles give identical ranges? Explain why using the complementary angle property.
- As the angle increases from 15° to 45°, does the time of flight increase or decrease? Explain.
In the angle-comparison table, which pair of angles gives the same range?
A projectile is launched from the top of a 50 m cliff at 18 m/s. Calculate the range at each of the following angles and determine which gives the maximum range. Take $g = 9.8\ \text{m/s}^2$.
Use the formula: $t_{\text{flight}} = \dfrac{v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh}}{g}$, then $R = v\cos\theta \times t_{\text{flight}}$.
| Angle $\theta$ | $v_y = v\sin\theta$ (m/s) | $\sqrt{v_y^2 + 2gh}$ (m/s) | $t_{\text{flight}}$ (s) | $v_x = v\cos\theta$ (m/s) | Range $R$ (m) |
|---|---|---|---|---|---|
| 20° | ________ | ________ | ________ | ________ | ________ |
| 35° | ________ | ________ | ________ | ________ | ________ |
| 45° | ________ | ________ | ________ | ________ | ________ |
| 55° | ________ | ________ | ________ | ________ | ________ |
Conclusion: What is the optimal launch angle from this cliff? Is it greater than, less than, or equal to 45°? Explain why in terms of horizontal velocity and flight time.
Misconceptions — final check
Copy into your books
Range Equation
- $R = v^2\sin(2\theta)/g$ (level ground only)
- $R_{\text{max}} = v^2/g$ at $\theta = 45°$
- Assumes level ground and no air resistance
Complementary Angles
- $\theta$ and $(90° - \theta)$ give same range
- Example: 30° = 60°, 25° = 65°
- Proof: $\sin(2\theta) = \sin(180° - 2\theta)$
Height Effects
- Launch above landing → optimal angle < 45°
- Launch below landing → optimal angle > 45°
Cliff Formula
- $t = (v\sin\theta + \sqrt{v^2\sin^2\theta + 2gh})/g$
- Then $R = v\cos\theta \times t_{\text{flight}}$
A projectile is launched from the top of a tall building. The angle for maximum range will be:
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.
AnalyseBand 4(3 marks) 1. Derive the range equation $R = v^2\sin(2\theta)/g$ starting from the equations of motion. State all assumptions.
1 mark: time of flight $t = 2v\sin\theta/g$ with working · 1 mark: substitution into horizontal equation · 1 mark: use of double-angle identity and assumptions stated
ApplyBand 4(3 marks) 2. A projectile is launched at 25 m/s on level ground. Calculate the range for launch angles of (a) 25°, (b) 45°, (c) 65°. Comment on your results. ($g = 9.8\ \text{m/s}^2$)
1 mark each for (a) and (b) with working · 1 mark for comment on complementary angles
EvaluateBand 6(4 marks) 3. Assess the claim that "a 45° launch angle is always optimal for maximum range in sporting contexts." Use specific examples to support your evaluation.
1 mark: identifies condition (level ground, no air resistance) · 1 mark: launch/landing height effect (cliff → <45°, uphill → >45°) · 1 mark: at least two specific sporting examples with approximate angles · 1 mark: reasoned conclusion
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 drawn from the lesson bank.
Short Answer — Model Answers
Q1 — Derivation (3 marks):
Start with horizontal motion: $s_x = v_x t = v\cos\theta \cdot t$.
For vertical motion on level ground, set $s_y = 0$:
$0 = v\sin\theta \cdot t_{\text{flight}} - \tfrac{1}{2}g \cdot t_{\text{flight}}^2$
Solving: $t_{\text{flight}} = 2v\sin\theta/g$ (1 mark)
Substitute into horizontal equation:
$R = v\cos\theta \times 2v\sin\theta/g = v^2(2\sin\theta\cos\theta)/g$ (1 mark)
Using $\sin(2\theta) = 2\sin\theta\cos\theta$: $R = v^2\sin(2\theta)/g$ (1 mark)
Assumptions: level ground (same launch and landing height), no air resistance, constant $g = 9.8\ \text{m/s}^2$, point projectile.
Q2 — Range at Different Angles (3 marks):
(a) At 25°: $R = (25)^2 \times \sin(50°)/9.8 = 625 \times 0.766/9.8 = \mathbf{48.9\ \text{m}}$ (1 mark)
(b) At 45°: $R = 625 \times \sin(90°)/9.8 = 625/9.8 = \mathbf{63.8\ \text{m}}$ (1 mark)
(c) At 65°: $R = 625 \times \sin(130°)/9.8 = 625 \times 0.766/9.8 = \mathbf{48.9\ \text{m}}$
25° and 65° give the same range because they are complementary angles ($25° + 65° = 90°$). 45° gives the maximum range on level ground. (1 mark for comment)
Q3 — Assessment of 45° in Sport (4 marks):
The claim is incorrect. A 45° launch angle is only optimal under idealised conditions (level ground, no air resistance, point projectile).
Launch and landing heights differ: In most sports, the projectile is launched from above ground level and lands at ground level. This makes the optimal angle less than 45°, as the extra fall time benefits a shallower trajectory. (1 mark)
Launch speed varies with angle: Athletes cannot achieve the same launch speed at all angles. For example, in the long jump, the optimal takeoff angle is approximately 20° because the athlete's horizontal speed from the run-up is dominant — attempting a steeper angle would drastically reduce horizontal velocity. (1 mark)
Specific examples:
Golf drivers use approximately 10° loft because the ball is on a tee and aerodynamic lift carries the ball much further than vacuum prediction. Shot put has an optimal angle of approximately 38°, limited by release height and biomechanical force. Javelin is thrown at approximately 35°, optimised for distance within legal parameters. (1 mark for at least two specific examples with approximate angles)
Conclusion: While 45° is a useful theoretical reference for level-ground projectile motion, it is rarely the optimal angle in sporting contexts due to differing launch/landing heights, biomechanical speed limitations, and aerodynamic effects. (1 mark for reasoned conclusion)
Five timed questions on launch angle analysis. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaPractise launch angle and range concepts while you play.
At the start you were asked about the 1918 German Krupp Paris Gun, which fired 94 kg shells at 130 km range at an elevation of 55° — not the 45° that the ideal range equation predicts as optimal.
Now you understand why: the range equation $R = v^2\sin(2\theta)/g$ assumes level ground, no air resistance, and no Earth rotation. At extreme ranges like 130 km, air resistance (which increases with speed and is greatest early in the trajectory) and the Coriolis effect from Earth's rotation both modify the optimal angle significantly above 45°. The 170 s flight time — calculated from the vertical motion equations — confirms the shell reached the stratosphere, where air density changes the drag profile further.
The Paris Gun illustrates that the ideal model is a starting point, not the full story. Has your understanding of the range equation changed?