Projectiles at an angle: time of flight, range and greatest height

Greatest height and time of flight

Take up as positive. With vertical component U sin α:

• at the greatest height the vertical velocity is 0, so 0 = (U sin α)² - 2gH gives H = U² sin² α / (2g)
• it returns to the launch level when s = 0: 0 = (U sin α)t - (1/2)gt², so the time of flight is T = 2U sin α / g.

Range on horizontal ground

Range R = (U cos α) × T = 2U² sin α cos α / g = U² sin 2α / g. The range is greatest when sin 2α = 1, that is α = 45°.

Pearson often asks you to show a formula like this. Write each step: the vertical equation, the time, then the horizontal distance. A quoted formula usually scores nothing on a show that question.

Launch from a height

If the particle lands below its starting point (for example from a cliff), use s = -h in s = (U sin α)t - 4.9t². This gives a quadratic; reject the negative root. To find how long the particle is above some level, solve for both times and subtract.

Worked example

A particle is projected from level ground at 24.5 m/s at angle α where tan α = 3/4. Find its greatest height, time of flight and range.

1. sin α = 0.6, cos α = 0.8, so the components are 19.6 m/s horizontally and 14.7 m/s vertically.
2. Greatest height: 0 = 14.7² - 19.6H, so H = 11.0 m (reached after 14.7/9.8 = 1.5 s).
3. Time of flight = 2 × 1.5 = 3 s.
4. Range = 19.6 × 3 = 58.8 m.

Answer: H = 11.0 m, T = 3 s, R = 58.8 m

Check yourself

1. A particle is projected from level ground at 20 m/s at 40° above the horizontal. Find its greatest height above the ground, to 3 significant figures.
   Show answer

   8.43 m. H = (20 sin 40°)² / 19.6 = 8.43 m.

2. A ball is kicked from level ground at 30 m/s at 25° above the horizontal. Find the time it is in the air, to 3 significant figures.
   Show answer

   2.59 s. 0 = (30 sin 25°)t - 4.9t², so t = 2(30 sin 25°)/9.8 = 2.59 s.

3. A particle is projected from level ground at 20 m/s at 30° above the horizontal. Find its range, to 3 significant figures.
   Show answer

   35.3 m. T = 2(10)/9.8 = 2.04 s; R = 20 cos 30° × T = 35.3 m.

4. A stone is projected at 14 m/s at 30° above the horizontal from a point 30 m above the sea. Find the time it takes to reach the sea, to 3 significant figures.
   Show answer

   3.29 s. Vertical component 7 m/s. Up positive: -30 = 7t - 4.9t², so 4.9t² - 7t - 30 = 0 and t = 3.29 s.

5. For the same stone (14 m/s at 30° from 30 m above the sea), find the horizontal distance from the foot of the cliff to where it lands, to 3 significant figures.
   Show answer

   39.9 m. 14 cos 30° × 3.29 = 39.9 m.

6. In the worked example (24.5 m/s with tan α = 3/4), find the speed of the particle 2 s after projection, to 3 significant figures.
   Show answer

   20.2 m/s. v_x = 19.6, v_y = 14.7 - 19.6 = -4.9, so speed = sqrt(19.6² + 4.9²) = 20.2 m/s.

7. A particle is projected with vertical component of velocity 14.7 m/s. For how long is it at least 9.8 m above its point of projection?
   Show answer

   1 s. 9.8 = 14.7t - 4.9t² gives t² - 3t + 2 = 0, so t = 1 or 2. It is above 9.8 m for 1 s.

8. For a fixed speed of projection from level ground, which angle gives the greatest range?
   • 30°
   • 60°
   • 45°
   • 90°
   Show answer

   45°. R = U² sin 2α / g is largest when sin 2α = 1, so α = 45°.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy

Last updated 8 November 2026. Spotted a mistake? Tell us.