Projectile motion — AQA A-Level Physical Education

Test yourself on Projectile motion with AQA A-Level practice questions.

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Factors affecting horizontal displacement of projectiles.

Projectile motion exam tips

Quick Revision Summary (Key Takeaway)

Projectile motion in AQA A-Level Physical Education is the study of objects launched into the air, moving under gravity alone. It combines horizontal uniform velocity with vertical uniform acceleration, and is assessed through calculations and analysis of sporting actions like a long jump or tennis serve.

Topic Overview

Projectile motion is a core topic in AQA A-Level Physical Education, covering the motion of objects projected into the air under the influence of gravity. It involves analysing horizontal and vertical components of motion independently, using equations of uniform acceleration. This topic is essential for understanding performance in sports such as athletics, football, and basketball, where the trajectory of a ball or body affects outcomes.

In the wider subject, projectile motion links to biomechanics and sports physics, helping students explain how factors like projection angle, speed, and height of release influence range and accuracy. It also underpins practical coaching points, such as adjusting technique to optimise distance or accuracy. Mastery of this topic supports both exam success and applied understanding in practical performance analysis.

Key Concepts

→Horizontal motion is constant velocity (no acceleration) because no horizontal force acts (ignoring air resistance).
→Vertical motion is uniformly accelerated due to gravity (g = 9.81 m/s^2 downwards), affecting time of flight and maximum height.
→The independence of horizontal and vertical components means calculations must be done separately and then combined.
→The range of a projectile is maximum at a projection angle of 45 degrees (for level ground), and complementary angles give equal ranges.
→Air resistance reduces both horizontal and vertical velocity, altering the trajectory and reducing range and maximum height.

Examiner Tips

💡Always resolve initial velocity into horizontal and vertical components before starting calculations. Show your working clearly to gain method marks.
💡Use the correct sign convention for vertical motion: take upwards as positive and downwards as negative (or vice versa) consistently.
💡When explaining the effect of air resistance, mention that it opposes motion, reducing both horizontal and vertical components, leading to a shorter range and lower maximum height.

Common Mistakes

Students often think horizontal velocity changes during flight. Correction: In the absence of air resistance, horizontal velocity remains constant because there is no horizontal force.
Students may believe that a larger projection angle always increases range. Correction: Range is maximum at 45 degrees; angles above or below reduce range, and complementary angles give the same range.
Students sometimes forget that vertical velocity at maximum height is zero. Correction: At the peak, vertical velocity is momentarily zero, but horizontal velocity is unchanged.

Revision Plan

1Day 1-2: Review vector resolution and kinematic equations. Practice resolving velocities at various angles.
2Day 3-4: Learn the independence of horizontal and vertical motion. Solve basic problems for time of flight, maximum height, and range.
3Day 5-6: Apply projectile motion to sporting examples (e.g., long jump, basketball shot). Analyse how angle, speed, and release height affect performance.
4Day 7-8: Complete exam-style questions, focusing on 4-6 mark explanations and calculations. Mark your work using the mark scheme.
5Day 9-10: Revise common misconceptions and examiner tips. Create flashcards for key formulas and definitions, and test yourself with active recall.

Exam Question Types

📋Calculation questions: Calculate range, maximum height, or time of flight from given initial velocity and angle. Advice: Always resolve components first and show all steps.
📋Explanation questions: Explain how changing projection angle affects range and maximum height. Advice: Use specific terminology like 'complementary angles' and 'vertical component'.
📋Analysis of sporting actions: Describe and explain the projectile motion in a given sport (e.g., a tennis serve). Advice: Link to horizontal and vertical components and mention air resistance if relevant.
📋Graph interpretation: Interpret velocity-time or displacement-time graphs for projectile motion. Advice: Remember horizontal velocity is constant, vertical velocity changes linearly.

Command Word Expectations (AQA)

Calculate

You must use mathematical formulas to find a numerical answer. Show all working, including formula, substitution, and units. Marks are awarded for correct method and final answer.

Explain

You must provide reasons or mechanisms behind a phenomenon. Use precise terminology (e.g., 'horizontal velocity is constant', 'vertical acceleration is 9.81 m/s^2'). Typically 2-4 marks, with one mark per correct point.

Evaluate

You must weigh up strengths and weaknesses or consider different factors and come to a conclusion. For projectile motion, this might involve comparing the effects of air resistance versus no air resistance, or the optimal angle for a given sport. Requires a balanced argument and a final judgement.

How Students Lose Marks (Examiner Pitfalls)

Pitfall: Students often assume that horizontal velocity changes during flight or that vertical velocity is constant, leading to incorrect calculations and loss of marks in 4-6 mark questions.

❌ Weak Answer (Loses Marks):The horizontal velocity decreases because of air resistance, so the ball slows down and lands shorter.

Example improved answer:In the absence of air resistance, horizontal velocity remains constant throughout the flight because there is no horizontal force acting on the projectile. Vertical velocity changes uniformly due to the acceleration of free fall (g = 9.81 m/s^2 downwards). Therefore, horizontal and vertical components must be treated independently.

Examiner Tip: Always state that horizontal motion is constant velocity and vertical motion is uniform acceleration. Use separate columns for horizontal and vertical calculations to avoid mixing them up.

Pitfall: When asked to explain the effect of projection angle on range, students frequently forget that complementary angles (e.g., 30 degrees and 60 degrees) give the same range but different flight times and maximum heights.

❌ Weak Answer (Loses Marks):A larger angle always gives a longer range because the object goes higher.

Example improved answer:For a given projection speed, the maximum range is achieved at 45 degrees. Complementary angles (such as 30 degrees and 60 degrees) produce the same horizontal range, but the higher angle gives a greater maximum height and longer flight time. This is because the vertical component of velocity is larger at higher angles, increasing time of flight, while the horizontal component is smaller, reducing horizontal distance per unit time.

Examiner Tip: Use the range equation R = (u^2 sin 2theta)/g to show that sin 2theta is the same for complementary angles. Always relate the angle to the vertical and horizontal velocity components.

Step-by-Step Worked Solutions

Question: A football is kicked with an initial velocity of 20 m/s at an angle of 30 degrees to the horizontal. Calculate the horizontal distance travelled (range) before it hits the ground. Assume no air resistance and g = 9.81 m/s^2.

1.Step 1: Identify given facts: initial speed u = 20 m/s, angle theta = 30 degrees, g = 9.81 m/s^2. Resolve initial velocity into horizontal and vertical components: ux = u cos theta = 20 cos 30 = 17.32 m/s, uy = u sin theta = 20 sin 30 = 10 m/s.
2.Step 2: Apply core rule: time of flight is determined by vertical motion. Use vertical displacement equation: s = uy t - 0.5 g t^2. Since the ball returns to the same height, s = 0, so 0 = 10 t - 0.5 * 9.81 * t^2. Factorise: t(10 - 4.905 t) = 0, giving t = 0 (start) or t = 10 / 4.905 = 2.04 s.
3.Step 3: Calculate horizontal range: range = ux * t = 17.32 * 2.04 = 35.33 m. State final conclusion with units: The horizontal distance travelled is approximately 35.3 m.

Final Answer: Range = 35.3 m (to 3 significant figures).

Question: A long jumper takes off with a velocity of 9.0 m/s at an angle of 25 degrees to the horizontal. Calculate the maximum height reached during the jump. Use g = 9.81 m/s^2.

1.Step 1: Identify given facts: u = 9.0 m/s, theta = 25 degrees, g = 9.81 m/s^2. Resolve vertical component: uy = u sin theta = 9.0 sin 25 = 3.80 m/s.
2.Step 2: Apply core rule: at maximum height, vertical velocity vy = 0. Use kinematic equation: vy^2 = uy^2 - 2 g s. Rearrange for s: s = (uy^2) / (2 g) = (3.80^2) / (2 * 9.81) = 14.44 / 19.62 = 0.736 m.
3.Step 3: State final conclusion with units: The maximum height reached is approximately 0.74 m.

Final Answer: Maximum height = 0.74 m (to 2 significant figures).

Active Recall Memory Test

What is the horizontal acceleration of a projectile in the absence of air resistance?

Key Fact: Zero, because no horizontal force acts on the projectile.

At what angle is the range of a projectile maximum on level ground?

Key Fact: 45 degrees.

What is the vertical velocity of a projectile at its maximum height?

Key Fact: 0 m/s.

State the equation for the range of a projectile on level ground.

Key Fact: R = (u^2 sin 2theta) / g.

Frequently Asked Questions

What is projectile motion in A-Level PE?

Projectile motion is the motion of an object thrown or projected into the air, subject only to gravity and air resistance. In A-Level PE, you study how horizontal and vertical components of motion combine to determine the path (trajectory) of a projectile. It is used to analyse sports like long jump, basketball, and tennis.

How do you calculate the range of a projectile?

To calculate range, first resolve the initial velocity into horizontal (ux = u cos theta) and vertical (uy = u sin theta) components. Then find the time of flight using vertical motion (e.g., t = 2uy/g for level ground). Finally, multiply horizontal velocity by time: range = ux * t. For non-level ground, use the vertical displacement equation to find time.

Why does a 45-degree angle give maximum range?

The range equation R = (u^2 sin 2theta)/g shows that range depends on sin 2theta. The maximum value of sin 2theta is 1, which occurs when 2theta = 90 degrees, so theta = 45 degrees. This assumes no air resistance and level ground. In real sports, air resistance and release height can alter the optimal angle.

What is the effect of air resistance on projectile motion?

Air resistance opposes motion, reducing both horizontal and vertical velocity components. This decreases the range and maximum height, and makes the trajectory asymmetric (steeper descent). It also means the optimal angle for maximum range is slightly less than 45 degrees. In exams, you must state whether air resistance is ignored or considered.

How do you resolve initial velocity into components?

Use trigonometry: horizontal component ux = u cos theta, vertical component uy = u sin theta, where u is initial speed and theta is the angle of projection. For example, if u = 20 m/s at 30 degrees, ux = 20 cos 30 = 17.32 m/s and uy = 20 sin 30 = 10 m/s. Always check your calculator is in degrees mode.

What are complementary angles in projectile motion?

Complementary angles are pairs of angles that add up to 90 degrees (e.g., 30 degrees and 60 degrees). For a given initial speed, they produce the same horizontal range but different maximum heights and flight times. The higher angle gives greater maximum height and longer flight time. This is because sin 2theta is the same for both angles.

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