A-Level · Mathematics · AQA · Mark scheme decoded
AQA A-Level Mathematics: Motion Under Gravity Using Vectors and Projectiles — mark scheme explained
The short answer
In AQA A-Level Mathematics, understanding the motion of objects under gravity in a vertical plane using vectors is crucial. This topic primarily deals with projectiles, which are objects that are launched into the air and move under the influence of gravity alone.
The question
A ball is projected from the ground with an initial velocity of 25 m/s at an angle of 30° to the horizontal. Calculate the maximum height reached by the ball and the range of the projectile.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For projectile motion questions, marks are typically awarded for correctly breaking down the initial velocity into components, using the correct equations of motion, and providing accurate calculations. Partial credit may be given for showing correct method steps even if the final answer is incorrect.
What the command words demand
- Calculate
- Perform the necessary mathematical operations to find a specific value.
- Determine
- Find or establish a particular value or quantity using given information.
- Derive
- Show how a formula or equation is obtained from first principles.
- Explain
- Provide a clear and detailed account of why something happens or how it works.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate approximately 5-7 minutes per question to ensure you have enough time to carefully break down the problem, use the correct formulas, and check your work.
- 1. Identify the given values: Initial velocity, v 0 = 25 m/s Projection angle, θ = 30° Acceleration due to gravity, g = 9.81 m/s 20 marks
- 2. Calculate the vertical component of the initial velocity: v y = v 0 sin(θ) = 25 × sin(30°) = 25 × 0.5 = 12.5 m/s1 mark
- 3. Calculate the maximum height using the formula: H = (v y 2 ) / (2g) = (12.5 2 ) / (2 × 9.81) ≈ 7.96 m2 marks
- 4. Calculate the horizontal component of the initial velocity: v x = v 0 cos(θ) = 25 × cos(30°) ≈ 21.65 m/s1 mark
- 5. Calculate the range using the formula: R = (v 0 2 sin(2θ)) / g = (25 2 sin(60°)) / 9.81 ≈ 55.23 m2 marks
Final answer: The maximum height reached by the ball is approximately 7.96 m , and the range of the projectile is approximately 55.23 m .
Work through every step correctly and you earn all 6 marks.
Another worked example
A stone is thrown from a cliff that is 100 m high with an initial velocity of 20 m/s at an angle of 45° to the horizontal. Calculate the time it takes for the stone to reach the ground and the distance from the base of the cliff where it lands.
- 1. Identify the given values: Initial velocity, v 0 = 20 m/s Projection angle, θ = 45° Height of the cliff, y 0 = 100 m Acceleration due to gravity, g = 9.81 m/s 20 marks
- 2. Calculate the vertical component of the initial velocity: v y = v 0 sin(θ) = 20 × sin(45°) ≈ 14.14 m/s1 mark
- 3. Use the equation of motion for vertical displacement to find the time: y = v y t - ½gt 2 -100 = 14.14t - 4.905t 2 Solve the quadratic equation: 4.905t 2 - 14.14t - 100 = 0 Using the quadratic formula, t ≈ 6.38 s4 marks
- 4. Calculate the horizontal component of the initial velocity: v x = v 0 cos(θ) = 20 × cos(45°) ≈ 14.14 m/s1 mark
- 5. Calculate the horizontal distance using the time found: x = v x t = 14.14 × 6.38 ≈ 90.27 m2 marks
Final answer: The stone takes approximately 6.38 s to reach the ground, and it lands approximately 90.27 m from the base of the cliff.
Work through every step correctly and you earn all 8 marks.
Common mistakes
Forgetting to break the initial velocity into horizontal and vertical components.
Why it happens: Students often overlook this step, leading to incorrect calculations of displacement and time.
Fix: Always start by breaking down the initial velocity using trigonometric functions: v x = v 0 cos(θ) and v y = v 0 sin(θ) .
Using the wrong value for acceleration due to gravity.
Why it happens: Students sometimes use 10 m/s 2 instead of 9.81 m/s 2 , leading to significant errors in calculations.
Fix: Always use g = 9.81 m/s 2 unless otherwise specified in the problem.
Forgetting that the horizontal component of velocity remains constant.
Why it happens: Students may incorrectly assume that the horizontal component changes over time, leading to incorrect range calculations.
Fix: Remember that v x is constant and use it directly in the equation for horizontal displacement: x = v x t .
Using the wrong formula for maximum height or range.
Why it happens: Students may mix up the formulas for maximum height and range, leading to incorrect answers.
Fix: Memorize the correct formulas: H = (v 0 2 sin 2 (θ)) / (2g) and R = (v 0 2 sin(2θ)) / g .
Forgetting to consider the initial height when calculating time of flight or range.
Why it happens: Students may overlook the initial height, leading to incorrect calculations for problems involving projectiles launched from a height.
Fix: Include the initial height in the vertical displacement equation: y = y 0 + v y t - ½gt 2 .
Incorrectly solving quadratic equations for time of flight.
Why it happens: Students may make algebraic errors when solving the quadratic equation, leading to incorrect values for time.
Fix: Use the quadratic formula carefully: t = (-b ± √(b 2 - 4ac)) / (2a) . Double-check your calculations.
Forgetting to check units and significant figures.
Why it happens: Students may overlook the importance of consistent units and appropriate significant figures, leading to incorrect answers.
Fix: Always ensure that all values are in consistent units (e.g., meters and seconds) and round your final answer to an appropriate number of significant figures.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Projectile Motion | Resolve the launch velocity into components, then find maximum height and horizontal range. | 6 |
| Projectile Motion | Resolve initial velocity and use equations of motion to find flight time and landing distance. | 8 |
| Projectile Motion | Resolve initial velocity into components and find time to maximum height and total flight time. | 6 |
| Projectile Motion | Resolve initial velocity into components, then find time to maximum height and horizontal range. | 6 |
| Total across these question types | 26 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.