Projectiles in vector form and the trajectory equation

Vector form

With i horizontal and j vertically up, a particle projected from O with velocity (pi + qj) m/s has acceleration -9.8j, so

• v = pi + (q - 9.8t)j
• r = pt i + (qt - 4.9t²)j.

This is just M1 vector suvat with a = -gj. The particle is moving at 45° below the horizontal when the j part of v equals minus the i part.

The trajectory equation

Eliminate t between x = (U cos α)t and y = (U sin α)t - (1/2)gt²:

y = x tan α - g x² / (2U² cos² α).

Using 1/cos² α = 1 + tan² α this becomes y = x tan α - g x² (1 + tan² α) / (2U²), which is a quadratic in tan α. If the particle must pass through a given point (x, y), solving this quadratic usually gives two possible angles. The path is a parabola.

Direction and perpendicular velocities

The direction of motion at time t is given by tan θ = v_y / v_x. Two velocities are perpendicular when their scalar product is zero, which you may need when asked when the particle is moving at right angles to its initial direction.

Worked example

A particle is projected from O with velocity (7i + 14j) m/s. Find the time when it is moving at 45° below the horizontal, and its position then.

1. v = 7i + (14 - 9.8t)j.
2. At 45° below the horizontal: 14 - 9.8t = -7, so t = 21/9.8 = 15/7 = 2.14 s (3 s.f.).
3. r = 7t i + (14t - 4.9t²)j = 15i + (30 - 22.5)j = 15i + 7.5j.

Answer: t = 2.14 s, position (15i + 7.5j) m

Check yourself

1. A particle is projected from O at 14 m/s at 45° above the horizontal. Use the trajectory equation to find its height when it is 10 m horizontally from O.
   Show answer

   5 m. y = 10 tan 45° - 9.8(100)/(2 × 196 × 0.5) = 10 - 5 = 5 m.

2. A particle is projected from O with velocity (6i + 8j) m/s. Find its greatest height above O, to 3 significant figures.
   Show answer

   3.27 m. Vertically: 0 = 64 - 19.6H, so H = 3.27 m.

3. For the same particle (6i + 8j m/s from O), find the angle between its direction of motion and the horizontal 1 s after projection, in degrees to 1 decimal place.
   Show answer

   16.7 °. v = 6i + (8 - 9.8)j = 6i - 1.8j, so the angle below the horizontal is arctan(1.8/6) = 16.7°.

4. For the same particle (6i + 8j m/s from O), find the time when it is moving perpendicular to its initial direction, to 3 significant figures.
   Show answer

   1.28 s. (6i + 8j)·(6i + (8 - 9.8t)j) = 0 gives 36 + 64 - 78.4t = 0, so t = 1.28 s.

5. A ball is projected from O at 20 m/s and passes through the point 20 m horizontally and 5 m vertically above O. Find the smaller possible angle of projection, in degrees to 1 decimal place.
   Show answer

   30.0 °. 5 = 20 tan α - 9.8(400)(1 + tan² α)/(2 × 400) gives 4.9 tan² α - 20 tan α + 9.9 = 0, so tan α = 0.576 or 3.51. The smaller angle is 30.0°.

6. A particle is projected from O with velocity (10i + 4.9j) m/s. Find its horizontal distance from O when it is next at the level of O.
   Show answer

   10 m. j part: 4.9t - 4.9t² = 0, so t = 1 s. i part: 10 × 1 = 10 m.

7. What shape is the path of a projectile in the M2 model?
   • A circle
   • A straight line
   • A hyperbola
   • A parabola
   Show answer

   A parabola. y is a quadratic function of x, so the path is a parabola.

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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.