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 Video by Zeeshan Zamurred Projectiles

Edexcel A Level Maths: 6.3 Projectiles (projection at any angle)Watch the full walkthrough 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

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.

v_{y} = 0 at max height; range = u cos θ \times T

Use symmetry and the horizontal speed.

1Vertical:

u_{y} = 20 sin 30° = 10

. Returns when

t = \frac{2 u _{y}}{g}

T = \frac{2 ( 10 )}{9.8} \approx 2.04

Answer

T \approx 2.04

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

Formula recap

T = \frac{2 u sin θ}{g}

Time of flight (level ground).

H = \frac{( u sin θ ) ^{2}}{2 g}

Maximum height.

R = u cos θ \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

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