Projectile motion
Resolve launch velocity into horizontal and vertical components, then model each component with constant acceleration.
Section Paper 3 · Further Mechanics · Topics 3.1–3.6
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Model a projectile as a particle under constant gravity and discuss limitations
- Use horizontal and vertical equations to find velocity, range and greatest height
- Derive and apply the Cartesian trajectory when launch data are unknown
Separate the independent components
For a launch speed u at angle θ above the horizontal, the initial components are u cos θ horizontally and u sin θ vertically. With air resistance neglected, horizontal acceleration is zero and vertical acceleration is −g.
Use x=u cosθ·t and y=u sinθ·t−½gt² from the launch point. The velocity components are u cosθ and u sinθ−gt. The model assumes a particle, uniform gravity and no air resistance; real drag and wind can change the path.
Eliminate time to obtain the trajectory
From x=u cosθ·t, t=x/(u cosθ). Substitute into y to get y=x tanθ−gx²/(2u²cos²θ). This is a parabola for a horizontal launch surface, provided cosθ≠0.
For a landing at launch height, set y=0 and ignore t=0 to obtain range R=u²sin(2θ)/g. The greatest height occurs when vertical velocity is zero; do not use the same-height range formula if the landing height differs.
Pause and explain
Which habit best checks your work on projectile motion?
Worked example
A projectile is launched at 20 m s⁻¹ at 30° above horizontal. Take g=10 m s⁻². Find its time to greatest height and its maximum height.
Show the worked solution
- The initial vertical speed is 20 sin30°=10 m s⁻¹.
- At greatest height, vertical velocity 10−10t is zero, so t=1 s.
- The rise is 10(1)−½(10)(1²)=5 m.
Answer It reaches greatest height after 1 s, at 5 m above launch level.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle is projected horizontally at 12 m s⁻¹ from a 20 m cliff. With g=10 m s⁻², find its flight time and horizontal distance before impact.
Give me a hint
Use vertical motion to find t, then use constant horizontal speed.
Compare my reasoning
- Vertically, 20=½(10)t², so t²=4 and t=2 s.
- Horizontally, x=12t.
- Substitute t=2.
Flight time 2 s; horizontal distance 24 m.
Look for these in your work
- Vertical equation selected
- Positive time chosen
- Horizontal displacement calculated
For the 20 m s⁻¹, 30° launch in the worked example, find the range on level ground using g=10 m s⁻².
Give me a hint
Use R=u²sin2θ/g only because the launch and landing heights match.
Compare my reasoning
- R=20² sin60°/10.
- sin60°=√3/2.
- Simplify the exact range.
R=20√3 m (about 34.6 m).
Look for these in your work
- Same-height condition stated
- Double angle used
- Exact and approximate values consistent
A projectile lands at a point below its launch level. Explain why setting y=0 in the launch-origin equation gives the wrong non-zero flight time.
Give me a hint
The vertical coordinate must represent displacement from launch level, not absolute height above the ground.
Compare my reasoning
- The equation y=u sinθ·t−½gt² measures vertical displacement from the launch point.
- At landing below launch, y is negative rather than zero.
- Use the actual vertical displacement to solve for t, then find horizontal travel.
Use the negative landing displacement; y=0 applies only when launch and landing levels are equal.
Look for these in your work
- Coordinate origin interpreted
- Landing displacement sign correct
- Range shortcut condition identified
Common mistakes
- Using the launch speed as the vertical component
- Giving horizontal acceleration a value of g
- Assuming landing height equals launch height without checking
What can you explain now?
What is the vertical acceleration of an ideal projectile when upward is positive?
Compare with the explanation
−g
Gravity acts downward throughout the flight.
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Make it stick
Why can horizontal and vertical motion be solved separately, and when is the level-ground range formula valid?
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