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6.4MechanicsStretch

Projectile Motion Formulae

General formulae for time of flight, maximum height, range and the equation of the path can be derived once, then applied directly. Knowing how they are derived helps you adapt them.

24 min Projectiles
Edexcel A Level Maths: 6.4 Projectile Motion FormulaeCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Derive the standard projectile formulae
  • Find the equation of the trajectory
  • Apply the formulae to problems
  • Know their assumptions
1

The trajectory equation

Eliminating tt from the component equations gives the path: y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta} — a parabola.

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}
Equation of the trajectory.
Example 1What shape is the path of a projectile?
1The trajectory equation is quadratic in x.
2So the path is a parabola.
Answer
A parabola.

Tip — The trajectory equation comes from substituting t = x/(u cos θ) into the vertical equation.

2

Assumptions

These formulae assume no air resistance, constant gravity, and a projectile modelled as a particle. They are idealisations of real motion.

Formula recap

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}
Trajectory (parabola).
R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}
Range (level ground).

Common mistakes to avoid

Forgetting the modelling assumptions.
No air resistance, particle model, constant g.
Misremembering the range formula.
R = u² sin 2θ / g on level ground.

Key takeaways

  • Trajectory: y = x tan θ − gx²/(2u²cos²θ) — a parabola.
  • Range on level ground: R = u² sin 2θ / g.
  • Assumes no air resistance, particle model, constant g.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. The path of a projectile is a:

    • parabola
    • circle
    • straight line
    • hyperbola
    Show answer and explanation

    Answer: parabola

    Quadratic trajectory.

  2. The range on level ground is:

    • u2sin⁡2θg\frac{u^2\sin 2\theta}{g}
    • u2sin⁡θg\frac{u^2\sin\theta}{g}
    • usin⁡2θg\frac{u\sin 2\theta}{g}
    • u2gu^2 g
    Show answer and explanation

    Answer: u2sin⁡2θg\frac{u^2\sin 2\theta}{g}

    Standard result.

  3. The trajectory equation is obtained by:

    • eliminating t
    • integrating
    • differentiating
    • adding components
    Show answer and explanation

    Answer: eliminating t

    Combine x and y equations.

  4. Which is an assumption of the model?

    • no air resistance
    • variable gravity
    • large object
    • friction included
    Show answer and explanation

    Answer: no air resistance

    Idealisation.

  5. Maximum range (fixed u) occurs at θ =

    • 45°45°
    • 30°30°
    • 60°60°
    • 90°90°
    Show answer and explanation

    Answer: 45°45°

    sin 2θ = 1.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 4 lessons

Put the methods into practice: see how A-Level maths duels work. An account is required to play; the overview and sample are public.