Projectile motion: splitting the motion and solving problems step by step
Projectile motion is the curved path of an object that is thrown or launched and then moves under gravity alone. The key idea is that the horizontal and vertical motions are independent: horizontally the velocity stays constant (ignoring air resistance), and vertically the object accelerates downwards at g. Solve each direction separately with the equations of uniform acceleration, linked by the same time.
Why the motion splits in two
Gravity acts only vertically, so it changes the vertical velocity but not the horizontal one. AQA's A-level Physics specification (3.4.1.4) describes this as the independent effect of motion in the horizontal and vertical directions of a uniform gravitational field, with problems solvable using the equations of uniform acceleration. In A-level Maths, AQA content Q5 covers projectiles as motion under gravity in a vertical plane.
| Horizontal | Vertical | |
|---|---|---|
| Acceleration | 0 | g downwards (often 9.81 m s⁻² in physics) |
| Initial velocity | u cos θ | u sin θ |
| Velocity | Constant | Changes by g every second |
| Equation to use | s = ut | suvat with a = −g (taking up as positive) |
Use the value of g that the question gives.
How to solve a projectile problem
- Resolve the launch velocity into horizontal (u cos θ) and vertical (u sin θ) components.
- Choose a positive direction for vertical motion, usually upwards, so a = −g.
- Use the vertical motion to find the time: for example, the time to reach the ground or the top.
- Use that time in the horizontal motion, s = ut, to find the distance.
- For the final velocity, find both components and combine them with Pythagoras and trigonometry.
At the top of the path the object stops rising but keeps moving sideways: only its upward velocity has fallen to nothing.
Worked examples
Example 1: rolling off a table
A marble rolls off a table 1.25 m high at 3.0 m s⁻¹. Find where it lands and how fast it is going. Take g = 9.81 m s⁻².
- Vertical: u = 0, s = 1.25 m, so 1.25 = ½ × 9.81 × t², giving t = 0.505 s.
- Horizontal: distance = 3.0 × 0.505 = 1.51 m from the table.
- Vertical speed on landing: v = gt = 9.81 × 0.505 = 4.95 m s⁻¹.
- Landing speed = √(3.0² + 4.95²) = 5.79 m s⁻¹, at tan⁻¹(4.95 ÷ 3.0) ≈ 58.8° below the horizontal.
Example 2: launched at an angle from the ground
A ball is thrown at 15 m s⁻¹, 30° above the horizontal, and lands at the same height.
- Time to the top: 0 = 7.50 − 9.81t, so t = 0.765 s. Time of flight is twice that, 1.53 s.
- Maximum height: 0 = 7.50² − 2 × 9.81 × s, so s = 2.87 m.
- Range: 12.99 × 1.53 = 19.9 m.
Here x is the horizontal distance and y is the height, both in metres. The axes are drawn to different scales, so the path looks steeper than it really is.
Example 3: launched upwards from a cliff
A stone is thrown at 12 m s⁻¹, 40° above the horizontal, from a cliff 10 m high. Vertically, taking up as positive, the stone lands at s = −10 m:
Take the positive root of the quadratic. The horizontal distance is 9.19 × 2.42 ≈ 22.2 m.
Try a projectile problem in our Laws of Motion course, then press Mark to check your answer in the course:
Find the flight time or the landing distance of a ball fired sideways off a cliff.
Press Mark to have your answer marked in the Laws of Motion course.
Air resistance
The equations above ignore air resistance. AQA's A-level Physics specification asks for a qualitative understanding of drag and lift forces and terminal speed. With air resistance:
- the horizontal velocity decreases, so the range is shorter
- the maximum height is lower
- the path is no longer a symmetrical parabola: the descent is steeper than the climb
Common mistakes
- Using the full launch speed in one direction. Resolve it into components first.
- Mixing up sin and cos. For an angle above the horizontal, the vertical component is u sin θ.
- Saying the velocity is zero at the top. Only the vertical component is zero.
- Inconsistent signs. If up is positive, g and downward displacements are negative.
- Taking the wrong root of a quadratic in t. Time must be positive.
Common questions
What is projectile motion?
Why are horizontal and vertical motion independent?
How do you find the time of flight?
How do you find the range of a projectile?
Is projectile motion in A-level Physics and Maths?
Revision cards
Say the answer first, then flip the card to check it.
Practise with a short timed paper
Projectile motion is in the first year of A-level Physics and in A-level Maths mechanics. The daily A-level Physics: Year 1 paper, and the The daily A-level Maths: Mechanics paper are short timed papers of 8 questions, with a new one every day. For the formulas in one place, see our projectile motion calculator guide.
Where to go next
- Solve it
Projectile Motion Simulator
Launch projectiles and read off range, height and flight time
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Laws of Motion
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A-level Physics: Year 1
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