AQA A-Level Physics Paper 1, June 2023: Question 18

1 mark · Medium difficulty · Multiple Choice

Calculate the magnitude of the displacement of a ball kicked at 45 degrees to the horizontal from point P when it reaches its maximum height after a time of 2.0 s.

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Question

Multiple choice question 18. A diagram shows a ball kicked from point P on level ground at 45 degrees to the horizontal, following a parabolic trajectory to its maximum height. Text states the ball reaches its maximum height after 2.0 s with air resistance ignored, and asks for the displacement of the ball from P when at its maximum height. Four options are provided: A 20 m, B 40 m, C 45 m, D 60 m.
Question text

18 A ball is kicked from point P on level ground. The ball initially travels at 45° to the

horizontal.

The ball reaches its maximum height after a time of 2.0 s.

Air resistance can be ignored.

What is the displacement of the ball from P when at its maximum height?

[1 mark]

A 20 m

B 40 m

C 45 m

D 60 m

Mark scheme

Show the mark scheme Mark scheme for question 18 showing the correct answer is C, corresponding to 45 m.

18 C 45 m

How to answer it

A-Level Physics Study Guide: Projectile Displacement

What this question tests

This question assesses your ability to analyze two-dimensional projectile motion by resolving vectors into horizontal and vertical components, applying SUVAT kinematic equations independently in both directions, and calculating resultant displacement using Pythagoras' theorem.

Question 18

Displacement of a Projectile at Maximum Height

✅ Correct Answer

Option C (45 m) is the correct choice.

💡 Key Knowledge

  • Horizontal and vertical motions are independent of one another.
  • At maximum height, vertical velocity ( v ) is 0 m s⁻¹ .
  • Resultant displacement ( s ) requires combining both horizontal displacement ( sₓ ) and vertical displacement ( sᵧ ) using s = √(sₓ² + sᵧ²) .

🧠 Exam Technique

  • Do not just calculate the vertical height or horizontal range alone; read carefully that the question asks for the resultant displacement from point P.
  • Use symmetry and components: since the launch angle is 45° , the initial horizontal and vertical velocity components are equal ( uₓ = uᵧ ).

❌ Common Errors

  • Only finding height: Calculating just vertical displacement ( sᵧ = 19.6 m ) or forgetting to add horizontal travel leads to wrong distractors.
  • Trig mix-ups: Incorrectly applying sine and cosine components for the 45° angle (though sin(45) equals cos(45)).

📐 Step-by-Step Calculation

  1. Find vertical displacement ( sᵧ ) at maximum height:
    Using SUVAT vertically where uᵧ = u sin(45) , v = 0 , a = -9.81 m s⁻² , and t = 2.0 s :
    sᵧ = ut + 0.5at²
    Alternatively, use average velocity: sᵧ = ((u + v) / 2) × t
    First find uᵧ using vertical deceleration: v = u + at → 0 = uᵧ - (9.81 × 2.0) → uᵧ = 19.62 m s⁻¹ .
    sᵧ = (19.62 / 2) × 2.0 = 19.62 m (or using 9.81 × 2.0² / 2 = 19.62 m ).
  2. Find horizontal velocity ( uₓ ):
    Since the launch angle is 45° , horizontal velocity equals vertical velocity: uₓ = uᵧ = 19.62 m s⁻¹ .
  3. Find horizontal displacement ( sₓ ) at t = 2.0 s :⁰
    sₓ = uₓ × t = 19.62 m s⁻¹ × 2.0 s = 39.24 m .
  4. Calculate resultant displacement ( s ):
    s = √(sₓ² + sᵧ²) = √(39.24² + 19.62²) = √(1539.78 + 384.94) = √1924.72 ≈ 43.87 m .
    Rounding to 2 significant figures (matching the data given in the question: 45° and 2.0 s ) yields 45 m (Option C).
Examiner Note: Top-level responses quickly recognized that for a 45° launch angle, horizontal displacement is always twice the vertical displacement at peak height ( sₓ = 2 sᵧ ), streamlining the calculation to s = √((2sᵧ)² + sᵧ²) = sᵧ√5 .

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.