Pearson Edexcel · A-Level · Mathematics
Course version: 9MA0 · Version from Mar 2026
Model motion under gravity in a vertical plane using vectors; projectiles.
Full specification wording
Kinematics: 7.5 Model motion under gravity in a vertical plane using vectors; projectiles.
A projectile moves in a vertical plane under gravity alone, so its acceleration is constant and directed downwards with magnitude g, approximately 9.8 m s⁻².
Checked against the Pearson Edexcel 9MA0 specification (opens in a new tab) on .
- Point 5 of 5
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
A projectile moves in a vertical plane under gravity alone, so its acceleration is constant and directed downwards with magnitude g, approximately 9.8 m s⁻². Taking horizontal and vertical axes, the horizontal acceleration is zero and the vertical acceleration is −g, giving r = r₀ + u t + ½a t² with a = −g ĵ. The horizontal velocity component u cos θ stays constant while the vertical component u sin θ − gt changes linearly. This yields time of flight from the vertical displacement equation, maximum height when the vertical velocity is zero, and horizontal range from the horizontal velocity multiplied by the time of flight. The model neglects air resistance and treats g as constant, so it applies only while the projectile is in free flight.
What earns the marks
- Resolves the initial velocity into horizontal and vertical components using u cos θ and u sin θ.
- Sets up the vector equation r = r₀ + u t + ½a t² with a = −g ĵ and applies it componentwise.
- Finds time of flight by solving the vertical displacement equation for the time when the projectile returns to the required height.
- Finds maximum height by setting the vertical velocity component to zero and substituting the resulting time.
- Finds horizontal range from the constant horizontal velocity multiplied by the total time of flight.
- States the modelling assumptions, including negligible air resistance and constant g, and comments on their effect.
Common mistakes and how to put them right
Using g as positive in the vertical equation when upwards has been taken as positive.
Put it right: Keep a = −g consistent with the chosen positive direction.
Assuming the horizontal velocity changes during flight.
Put it right: Treat the horizontal component as constant because horizontal acceleration is zero.
Confusing time to maximum height with total time of flight.
Put it right: Note that for level ground the total time is twice the time to maximum height.
Exam tips
- Draw a diagram with axes and mark the positive direction before writing any equations.
- Keep horizontal and vertical equations separate and label each with its component to avoid mixing them.
- Check whether the projectile lands at the same height as it was launched, since this changes the time-of-flight calculation.
Try the lesson
The start of the Kinematics: 7.5 Model motion under gravity in a vertical plane using vectors; projectiles. lesson, made from checked teaching notes for this specification point.
Have a go first
A ball is projected horizontally at 8 m s⁻¹ from the top of a cliff and falls freely under gravity. Taking g = 9.8 m s⁻², what is its vertical speed after 3 seconds?
Listen
A projectile moves in a vertical plane under gravity alone, so its acceleration is constant and directed downwards with magnitude g, approximately 9.8 m s⁻². Taking horizontal and vertical axes, the horizontal acceleration is zero and the vertical acceleration is −g, giving r = r₀ + u t + ½a t² with a = −g ĵ. The horizontal velocity component u cos θ stays constant while the vertical component u sin θ − gt changes linearly. This yields time of flight from the vertical displacement equation, maximum height when the vertical velocity is zero, and horizontal range from the horizontal velocity multiplied by the time of flight. The model neglects air resistance and treats g as constant, so it applies only while the projectile is in free flight. A strong answer earns marks for these points. Resolves the initial velocity into horizontal and vertical components using u cos θ and u sin θ. Sets up the vector equation r = r₀ + u t + ½a t² with a = −g ĵ and applies it componentwise. Finds time of flight by solving the vertical displacement equation for the time when the projectile returns to the required height. Finds maximum height by setting the vertical velocity component to zero and substituting the resulting time. A common slip: Using g as positive in the vertical equation when upwards has been taken as positive. Instead, keep a = −g consistent with the chosen positive direction. Exam tip: Draw a diagram with axes and mark the positive direction before writing any equations.
Key words
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Independence of horizontal and vertical motion
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Constant downward acceleration under gravity
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Vector modelling of motion in a vertical plane
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Idealised assumptions and their limitations
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See every definition
- Independence of horizontal and vertical motion
- The horizontal acceleration is zero so the horizontal velocity component stays constant, while the vertical component changes linearly.
- Constant downward acceleration under gravity
- Acceleration is constant and directed downwards with magnitude g, approximately 9.8 m s⁻², giving a = −g ĵ.
- Vector modelling of motion in a vertical plane
- Using r = r₀ + u t + ½a t² with a = −g ĵ and applying it componentwise to motion in a vertical plane.
- Idealised assumptions and their limitations
- The model neglects air resistance and treats g as constant, so it applies only while the projectile is in free flight.
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