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6.3MechanicsStretch

Projection at Any Angle

A full projectile problem combines resolved components with suvat. You can find the time of flight, the maximum height, and the range for a projectile launched at any angle.

26 min Projectiles
Edexcel A Level Maths: 6.3 Projectiles (projection at any angle)Curated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Find the time of flight
  • Find the maximum height
  • Find the horizontal range
  • Use symmetry of the path
1

Key results

At maximum height the vertical velocity is zero. The path is symmetric, so the time up equals the time down. Total time of flight comes from the vertical motion returning to its launch level.

vy=0 at max height;range=ucos⁡θ×Tv_y = 0 \text{ at max height}; \quad \text{range} = u\cos\theta \times T
Use symmetry and the horizontal speed.
Example 1A ball launched at 20 m/s at 30° on level ground. Find the time of flight (g = 9.8).
1Vertical: uy=20sin⁡30°=10u_y = 20\sin30° = 10. Returns when t=2uygt = \tfrac{2u_y}{g}.
2T=2(10)9.8≈2.04T = \tfrac{2(10)}{9.8} \approx 2.04 s.
Answer
T≈2.04T \approx 2.04 s

Tip — Time of flight on level ground is 2u sin θ / g; max height is at half that time.

Formula recap

T=2usin⁡θgT = \frac{2u\sin\theta}{g}
Time of flight (level ground).
H=(usin⁡θ)22gH = \frac{(u\sin\theta)^2}{2g}
Maximum height.
R=ucos⁡θ×TR = u\cos\theta \times T
Range.

Common mistakes to avoid

Using the full speed u in the vertical suvat.
Use the vertical component u sin θ.
Forgetting vertical velocity is zero at the top.
v_y = 0 at maximum height.

Key takeaways

  • At max height v_y = 0; the path is symmetric.
  • Time of flight (level) = 2u sin θ / g.
  • Range = horizontal speed × time of flight.

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. At maximum height the vertical velocity is:

    • 00
    • usin⁡θu\sin\theta
    • uu
    • gg
    Show answer and explanation

    Answer: 00

    Top of the path.

  2. The time of flight on level ground is:

    • 2usin⁡θg\frac{2u\sin\theta}{g}
    • usin⁡θg\frac{u\sin\theta}{g}
    • 2ucos⁡θg\frac{2u\cos\theta}{g}
    • ug\frac{u}{g}
    Show answer and explanation

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

    Up and down.

  3. The range equals:

    • ucos⁡θ×Tu\cos\theta \times T
    • usin⁡θ×Tu\sin\theta \times T
    • u×Tu \times T
    • 12gT2\tfrac12 g T^2
    Show answer and explanation

    Answer: ucos⁡θ×Tu\cos\theta \times T

    Horizontal speed × time.

  4. For 20 m/s at 30° on level ground, T ≈ (g=9.8):

    • 2.042.04 s
    • 1.021.02 s
    • 44 s
    • 1010 s
    Show answer and explanation

    Answer: 2.042.04 s

    2(10)/9.8.

  5. The maximum height occurs at:

    • half the time of flight
    • the start
    • the end
    • twice the time of flight
    Show answer and explanation

    Answer: half the time of flight

    By symmetry.

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.