Projectiles: the model and horizontal projection

The model

• The object is a particle (no size, no spin).
• There is no air resistance, so the only force is its weight.
• g = 9.8 m/s² is constant and acts vertically downwards.

So the horizontal component of velocity never changes, and the vertical motion obeys the suvat equations with a = -9.8 (taking up as positive).

Resolving the initial velocity

A particle projected with speed U at angle α above the horizontal has:

• horizontal component U cos α (constant throughout)
• vertical component U sin α (changes at 9.8 m/s every second).

The link between the two directions is time: find t from one direction and use it in the other.

Horizontal projection

If a particle is thrown horizontally from a height h, its initial vertical velocity is 0. The time to land comes from h = (1/2)gt² (taking down as positive). The horizontal distance is then (horizontal speed) × t.

To find the speed at any moment, combine the components: speed = sqrt(v_x² + v_y²). The direction is the angle below the horizontal, tan θ = v_y / v_x.

Worked example

A ball is thrown horizontally at 12 m/s from the top of a cliff 19.6 m high. Find the time to reach the ground, the horizontal distance travelled and the speed on landing.

1. Vertically (down positive): 19.6 = 0 + 4.9t², so t² = 4 and t = 2 s.
2. Horizontally: distance = 12 × 2 = 24 m.
3. Vertical velocity on landing: 0 + 9.8 × 2 = 19.6 m/s.
4. Speed = sqrt(12² + 19.6²) = 23.0 m/s, at 58.5° below the horizontal.

Answer: 2 s, 24 m, 23.0 m/s

Check yourself

1. A particle is projected at 25 m/s at 30° above the horizontal. Find the horizontal component of its velocity, to 3 significant figures.
   Show answer

   21.7 m/s. 25 cos 30° = 21.7 m/s.

2. A stone is thrown horizontally at 15 m/s from a point 44.1 m above level ground. Find the horizontal distance it travels before landing.
   Show answer

   45 m. Vertically: 44.1 = 4.9t², so t² = 9 and t = 3 s. Horizontally: 15 × 3 = 45 m.

3. A ball is thrown horizontally from a height of 10 m. Find the time it takes to reach the ground, to 3 significant figures.
   Show answer

   1.43 s. 10 = 4.9t², so t = sqrt(10/4.9) = 1.43 s. The horizontal speed does not affect this.

4. A particle is projected horizontally. Find the size of its vertical velocity 1.5 s later.
   Show answer

   14.7 m/s. v = 0 + 9.8 × 1.5 = 14.7 m/s downwards.

5. A particle is projected horizontally at 10 m/s. Find the angle its velocity makes with the horizontal 2 s later, in degrees to 1 decimal place.
   Show answer

   63.0 °. v_x = 10, v_y = 19.6, so tan θ = 19.6/10 and θ = 63.0° below the horizontal.

6. Which quantity stays constant throughout the flight of a projectile in the M2 model?
   • The vertical component of velocity
   • The speed
   • The horizontal component of velocity
   • The direction of motion
   Show answer

   The horizontal component of velocity. With no air resistance there is no horizontal force, so the horizontal component of velocity is constant.

7. Which modelling assumption makes the horizontal velocity of a projectile constant?
   • Air resistance is ignored
   • The particle has no size
   • g is taken as 9.8 m/s²
   • The ground is horizontal
   Show answer

   Air resistance is ignored. Air resistance would be a horizontal force slowing the particle; ignoring it leaves no horizontal force.

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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.