AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Problems will be solvable using the equations of uniform acceleration.
Full specification wording
Projectile motion: Problems will be solvable using the equations of uniform acceleration.
Projectile problems can be solved by applying the equations of uniform acceleration separately to the horizontal and vertical directions.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 2 of 7
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Projectile problems can be solved by applying the equations of uniform acceleration separately to the horizontal and vertical directions. Horizontally, acceleration is zero, so displacement equals horizontal velocity multiplied by time. Vertically, acceleration is g downwards, so the standard equations link initial vertical velocity, final vertical velocity, displacement, acceleration and time. A reliable method is to list the known quantities in each direction, choose the equation containing the unknown, and solve.
For example: A ball launched at 20 m s⁻¹ at 30° above the horizontal has initial vertical velocity 10 m s⁻¹ and horizontal velocity about 17 m s⁻¹; the vertical equation gives the time to reach maximum height, and the horizontal equation then gives the horizontal distance travelled in that time.
What earns the marks
- Select the appropriate equation of uniform acceleration for the direction being considered, using the known and unknown quantities.
- Apply v = u + at, s = ut + ½at², v² = u² + 2as or s = ½(u + v)t correctly to horizontal or vertical motion.
- Use a = 0 horizontally and a = g downwards vertically, taking care with the sign of g.
- Recognise that the time of flight is common to both directions and use it to link horizontal and vertical calculations.
- Solve for quantities such as maximum height, time of flight or horizontal range by substituting values with consistent units.
Common mistakes and how to put them right
Using the resultant initial speed instead of the resolved component in a vertical or horizontal equation.
Put it right by resolving the velocity first.
Mixing horizontal and vertical quantities in one equation.
Put it right by keeping the two directions in separate columns.
Forgetting that at maximum height the vertical velocity is zero while the horizontal velocity is unchanged.
Put it right by applying v = 0 only to the vertical direction.
Exam tips
- Write down the equation you are using before substituting numbers, so the examiner can follow your reasoning.
- Use consistent signs for upward and downward quantities, and state which direction is positive.
- Round only at the end of a calculation and give the unit with every answer.
Try the lesson
The start of the Projectile motion: Problems will be solvable using the equations of uniform acceleration. lesson, made from checked teaching notes for this specification point.
Have a go first
A ball is dropped from rest and falls freely under gravity. Taking g = 9.8 m s⁻², what is its speed after 2.0 s?
Listen
Projectile problems can be solved by applying the equations of uniform acceleration separately to the horizontal and vertical directions. Horizontally, acceleration is zero, so displacement equals horizontal velocity multiplied by time. Vertically, acceleration is g downwards, so the standard equations link initial vertical velocity, final vertical velocity, displacement, acceleration and time. A reliable method is to list the known quantities in each direction, choose the equation containing the unknown, and solve. For example, a ball launched at 20 m s⁻¹ at 30° above the horizontal has initial vertical velocity 10 m s⁻¹ and horizontal velocity about 17 m s⁻¹; the vertical equation gives the time to reach maximum height, and the horizontal equation then gives the horizontal distance travelled in that time. A strong answer earns marks for these points. Select the appropriate equation of uniform acceleration for the direction being considered, using the known and unknown quantities. Apply v = u + at, s = ut + ½at², v² = u² + 2as or s = ½(u + v)t correctly to horizontal or vertical motion. Use a = 0 horizontally and a = g downwards vertically, taking care with the sign of g. Recognise that the time of flight is common to both directions and use it to link horizontal and vertical calculations. A common slip: Using the resultant initial speed instead of the resolved component in a vertical or horizontal equation. Instead, resolving the velocity first. Exam tip: Write down the equation you are using before substituting numbers, so the examiner can follow your reasoning.
See it
Key words
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Equations of uniform acceleration
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Common time of flight
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Sign conventions in projectile calculations
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See every definition
- Equations of uniform acceleration
- Problems will be solvable using these, applying them separately to the horizontal and vertical directions.
- Common time of flight
- The time is common to both directions and can be used to link horizontal and vertical calculations.
- Sign conventions in projectile calculations
- Consistent signs are used for upward and downward quantities, taking care with the sign of g.
Label the diagram
Some boxes have lost their names. Pick the right name for each number.
1Box 1
2Box 2
3Box 3
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