AQA A-Level Mathematics: Motion Under Gravity Using Vectors and Projectiles — mark scheme explained

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

What each mark is really for — in plain English — and the wording trap that loses it.

• S1

  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 2

• S2

  2. Calculate the vertical component of the initial velocity: v y = v 0 sin(θ) = 25 × sin(30°) = 25 × 0.5 = 12.5 m/s

• S3

  3. Calculate the maximum height using the formula: H = (v y 2 ) / (2g) = (12.5 2 ) / (2 × 9.81) ≈ 7.96 m

• S4

  4. Calculate the horizontal component of the initial velocity: v x = v 0 cos(θ) = 25 × cos(30°) ≈ 21.65 m/s

• S5

  5. Calculate the range using the formula: R = (v 0 2 sin(2θ)) / g = (25 2 sin(60°)) / 9.81 ≈ 55.23 m

Model answer

Worked through, with each step tagged to the mark it earns.

1. S1

   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 2

2. S2

   2. Calculate the vertical component of the initial velocity: v y = v 0 sin(θ) = 25 × sin(30°) = 25 × 0.5 = 12.5 m/s

3. S3

   3. Calculate the maximum height using the formula: H = (v y 2 ) / (2g) = (12.5 2 ) / (2 × 9.81) ≈ 7.96 m

4. S4

   4. Calculate the horizontal component of the initial velocity: v x = v 0 cos(θ) = 25 × cos(30°) ≈ 21.65 m/s

5. S5

   5. Calculate the range using the formula: R = (v 0 2 sin(2θ)) / g = (25 2 sin(60°)) / 9.81 ≈ 55.23 m

6. 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 .

Common mistakes

• Forgetting to break the initial velocity into horizontal and vertical components. — 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. — Always use g = 9.81 m/s 2 unless otherwise specified in the problem.
• Forgetting that the horizontal component of velocity remains constant. — 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. — 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. — 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. — Use the quadratic formula carefully: t = (-b ± √(b 2 - 4ac)) / (2a) . Double-check your calculations.
• Forgetting to check units and significant figures. — 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

• Full worked solution (all marking points)6 marks

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