Calculating Projectile Motion Revision Notes | A Level AQA Physics
Find time of flight from vertical motion, then range = horizontal velocity × time, as horizontal velocity is constant. Free AQA notes and 12 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Calculating Projectile Motion is part of Force, energy and momentum in A Level AQA Physics.
6 min
Reading time
14
Lessons in this topic
A Level AQA
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What this lesson covers
Find time of flight from vertical motion, then range = horizontal velocity × time, as horizontal velocity is constant. Free AQA notes and 12 questions answered.
Explain why horizontal and vertical components of projectile motion are treated independently
Apply the equations of uniform acceleration to both components of projectile motion
Calculate the range, time of flight, and maximum height of a projectile
Analyse projectile problems by resolving initial velocity into horizontal and vertical components
Key ideas to keep in view
Use this lesson to stay inside Force, energy and momentum while connecting the detail back to the wider Mechanics and materials topic.
Projectile motion: Motion under gravity alone, where horizontal velocity remains constant and vertical motion undergoes uniform acceleration due to gravity. Applies when air resistance is negligible.
Independence of motion: The principle that horizontal and vertical components of a projectile's velocity and displacement do not affect each other and can be analysed separately using the equations of uniform acceleration.
Equations of uniform acceleration (suvat): A set of five kinematic equations linking displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) for motion with constant acceleration.
Range: The horizontal distance travelled by a projectile from launch to landing, calculated using the horizontal component of velocity multiplied by the total time of flight.
Revision notes
Read the core explanations from Calculating Projectile Motion before testing yourself from memory.
1. Why Projectile Motion Is Simpler Than It Looks
Picture a ball kicked horizontally off a cliff. Horizontally, nothing is pushing or pulling it — so it keeps moving at the same speed. Vertically, gravity pulls it downward with a constant acceleration of $9.8 \text{ m s}^{-2}$. These two things happen at the same time, but they do not interfere with each other.
That single insight — that horizontal and vertical motions are independent — is the key that unlocks every projectile problem on the AQA paper. Once you split the motion into two separate one-dimensional problems, each one is solved with the familiar equations of uniform acceleration (the suvat equations).
This independence is not an approximation — it follows directly from Newton's second law. Gravity acts only downward, so it only accelerates the object downward. There is no horizontal force, so there is no horizontal acceleration. The two axes are decoupled.
2. The Toolkit: Suvat Equations and How to Apply Them
Before tackling a projectile, you need the five equations of uniform acceleration firmly in mind. Each equation links four of the five quantities $s$, $u$, $v$, $a$, $t$ — so you choose whichever equation contains your three knowns and one unknown.
For the horizontal direction, acceleration $a = 0$, so every suvat equation simplifies. The most useful result is $s_x = u_x t$, where $u_x$ is the constant horizontal velocity. For the vertical direction, $a = g = 9.8 \text{ m s}^{-2}$ (downward), and you apply whichever suvat equation suits the unknowns in the question.
Choosing a consistent sign convention before you start is non-negotiable. The most reliable approach: take upward as positive and downward as negative. Then $a = -9.8 \text{ m s}^{-2}$ throughout, and any displacement below the launch point will come out negative automatically.
3. Resolving Initial Velocity into Components
When a projectile is launched at an angle $\theta$ above the horizontal with speed $u$, you cannot apply suvat directly — you first need to split $u$ into its horizontal and vertical parts. This is the same vector resolution you used in forces, applied here to velocity.
Once you have $u_x$ and $u_y$, treat them as the starting points for two completely separate suvat calculations. The horizontal calculation uses $a = 0$; the vertical calculation uses $a = -9.8 \text{ m s}^{-2}$ (taking upward as positive). The one quantity shared between both calculations is time — the projectile spends the same time in the air whether you are thinking about its horizontal or vertical journey.
Time is the bridge between the two directions. Solve for $t$ in the vertical direction first (using the vertical suvat equations), then substitute that same $t$ into the horizontal equation to find range. This two-step strategy works for virtually every projectile question.
4. Worked Example: Horizontal Launch from a Height
The simplest projectile scenario is a horizontal launch — the initial vertical velocity is zero, which removes one variable immediately. Work through this carefully; the method here is the template for all projectile problems.
Worked example: Horizontal Launch: Finding Time of Flight and Range
A ball is kicked horizontally at $15 \text{ m s}^{-1}$ from the edge of a cliff that is $45 \text{ m}$ high. Calculate (a) the time taken to reach the ground and (b) the horizontal range. Take $g = 9.8 \text{ m s}^{-2}$.
Take downward as positive.
Vertical: $u_y = 0$, $a = 9.8 \text{ m s}^{-2}$, $s_y = 45 \text{ m}$, $t =?$
Horizontal: $u_x = 15 \text{ m s}^{-1}$, $a = 0$ — Because the launch is horizontal, the initial vertical velocity is zero. Taking downward as positive here makes the vertical displacement and acceleration both positive, which keeps the arithmetic clean for this problem.
$s = ut + \frac{1}{2}at^2$
$45 = (0)t + \frac{1}{2}(9.8)t^2$
$45 = 4.9t^2$ — We know $s$, $u$, and $a$ vertically, and we want $t$. The equation $s = ut + \frac{1}{2}at^2$ contains exactly these four quantities. With $u_y = 0$, the $ut$ term vanishes.
$t^2 = \frac{45}{4.9} = 9.1837...$
$t = \sqrt{9.1837} = 3.03 \text{ s}$ (3 s.f.) — Dividing 45 by 4.9 gives 9.1837. Taking the square root: $\sqrt{9.1837} \approx 3.03 \text{ s}$. Always take the positive root — time cannot be negative. [A1]
$s_x = u_x \times t = 15 \times 3.03 = 45.5 \text{ m}$ (3 s.f.) — Horizontal acceleration is zero, so $s_x = u_x t$. Substituting $u_x = 15 \text{ m s}^{-1}$ and $t = 3.03 \text{ s}$ gives the range. [A1]
At $t = 3.03$ s, vertical drop: $\frac{1}{2} \times 9.8 \times 3.03 \times 3.03 = 4.9 \times 9.18 \approx 45 \text{ m}$ ✓ — Substituting $t$ back into the vertical equation confirms the 45 m drop. The horizontal range of about 45 m for a 15 m s⁻¹ horizontal speed over ~3 s is physically reasonable.
Angled launches add one extra step at the start — resolving the velocity — but the core strategy is identical. Solve vertically for time, then use that time horizontally. Here is a full walkthrough of a typical exam question.
Worked example: Angled Launch: Maximum Height and Range
A ball is launched from ground level at $20 \text{ m s}^{-1}$ at an angle of $30°$ above the horizontal. Calculate (a) the maximum height reached and (b) the total horizontal range. Take $g = 9.8 \text{ m s}^{-2}$.
$u_x = 20\cos 30° = 20 \times 0.866 = 17.3 \text{ m s}^{-1}$
$u_y = 20\sin 30° = 20 \times 0.500 = 10.0 \text{ m s}^{-1}$ (upward) — Always resolve first. $\cos 30° = 0.866$ and $\sin 30° = 0.500$ are exact enough for 3 s.f. working. Take upward as positive throughout. [M1]
$v^2 = u^2 + 2as$
0 = (10.0)^2 + 2(-9.8)(s)$
0 = 100 - 19.6s$
$s = \frac{100}{19.6} = 5.10 \text{ m}$ (3 s.f.) — At maximum height, the vertical velocity $v = 0$. Using $v^2 = u^2 + 2as$ with $u_y = 10.0 \text{ m s}^{-1}$ and $a = -9.8 \text{ m s}^{-2}$: $(10.0)^2 = 10.0 \times 10.0 = 100$. Then $s = 100 \div 19.6 = 5.10 \text{ m}$. [M1][A1]
$s = u_y t + \frac{1}{2}at^2$
$0 = 10.0t + \frac{1}{2}(-9.8)t^2$
$0 = 10.0t - 4.9t^2$
$0 = t(10.0 - 4.9t)$
$t = 0$ (launch) or $t = \frac{10.0}{4.9} = 2.04 \text{ s}$ — Setting $s = 0$ gives two solutions: $t = 0$ (the moment of launch) and $t = 2.04$ s (landing). We take the non-zero solution. Note: $10.0 \div 4.9 = 2.0408... \approx 2.04$ s. [M1]
Time to max height: $v = u + at \Rightarrow 0 = 10.0 - 9.8t \Rightarrow t = 1.02 \text{ s}$
Total flight time $= 2 \times 1.02 = 2.04 \text{ s}$ ✓ — For a symmetric trajectory (launch and landing at the same height), the time to maximum height is exactly half the total flight time. This confirms our answer of 2.04 s.
Answer: (a) Maximum height $= 5.10 \text{ m}$ (b) Range $= 35.3 \text{ m}$
6. Visualising the Trajectory
The path of a projectile is a parabola. This shape emerges directly from the mathematics: horizontal displacement grows linearly with time ($s_x = u_x t$), while vertical displacement grows with $t^2$ ($s_y = u_y t - \frac{1}{2}gt^2$). Combining these gives the characteristic curved path.
The symmetry of the parabola (when launch and landing heights are equal) is a useful check: the time to reach maximum height equals the time to fall back down, and the horizontal velocity component is the same magnitude at launch and landing. Examiners sometimes ask you to use this symmetry to simplify a calculation.
7. Assumptions and Limitations
Every projectile calculation in AQA A Level rests on two key assumptions. First, the gravitational field is uniform — $g$ is constant at $9.8 \text{ m s}^{-2}$ throughout the flight. Second, air resistance is negligible. Both assumptions are stated or implied in every exam question that asks you to use the suvat approach.
In reality, air resistance acts opposite to the direction of motion and reduces both horizontal and vertical components of velocity. This makes real trajectories non-parabolic and reduces the range compared to the idealised calculation. At A Level, you are expected to state these assumptions when asked to evaluate the model, and to recognise that real projectiles travel shorter distances than the equations predict.
8. Exam Technique and Common Mistakes
Projectile questions are among the most reliably structured on the AQA paper. The mark scheme almost always awards one method mark for resolving velocity correctly, one for selecting and applying the correct suvat equation, and one accuracy mark for the final answer with correct units. Losing marks here is nearly always due to one of a small number of recurring errors.
When a question uses the command word calculate, the mark scheme expects: the correct formula stated, values substituted with units, and a final answer with units to an appropriate number of significant figures (match the data given, typically 2–3 s.f.). When asked to explain why the two components are independent, use Newton's second law explicitly: gravity provides no horizontal force, so horizontal acceleration is zero, so horizontal velocity cannot change.
The independence of horizontal and vertical motion in a uniform gravitational field — and the ability to apply the suvat equations separately to each direction — is the single specification requirement underpinning every projectile calculation at A Level. Master the two-step strategy (resolve, then solve each direction with suvat), and you have a reliable method for any variant the examiner can set.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
How do you find the maximum height of a projectile?
To find maximum height, use v² = u² + 2as on the vertical component, with v = 0 at the top. For a ball launched at 20 m s⁻¹ at 30° to the horizontal, the vertical component is 20 sin 30° = 10.0 m s⁻¹, so 0 = 100 - 19.6s and the maximum height is 5.10 m.
How do you find the time of flight of a projectile?
The time of flight of a projectile comes from the vertical motion alone. For a ball launched from ground level with a vertical velocity of 10.0 m s⁻¹, setting s = 0 in s = ut + ½at² gives 0 = 10.0t − 4.9t², so t = 0 at launch or t = 2.04 s at landing.
Why are horizontal and vertical motion independent in projectiles?
Horizontal and vertical motion of a projectile are independent because gravity acts only downward, so it only accelerates the object vertically. With air resistance neglected there is no horizontal force, so horizontal velocity stays constant, while vertical velocity changes at g = 9.8 m s⁻². This follows directly from Newton's second law.
How do you calculate the range of a projectile?
The range of a projectile is found by solving the vertical motion for the time of flight, then using range = horizontal velocity × time, since the horizontal velocity is constant. A ball kicked horizontally at 15 m s⁻¹ from a 45 m cliff takes t = √(45 ÷ 4.9) = 3.03 s to land, so its range is 15 × 3.03 = 45.5 m.
How do you resolve the velocity of a projectile launched at an angle?
For a projectile launched at speed u and angle θ to the horizontal, the horizontal component is u cos θ and the vertical component is u sin θ. At 20 m s⁻¹ and 30°, these are 17.3 m s⁻¹ horizontally and 10.0 m s⁻¹ vertically. Each component is then solved separately with suvat, linked by the shared time.
What assumptions are made in projectile motion?
Projectile calculations assume the gravitational field is uniform, so g is constant at 9.8 m s⁻², and that air resistance is negligible. In reality air resistance acts opposite to the direction of motion and reduces both velocity components, so real trajectories are non-parabolic and real projectiles travel shorter distances than the equations predict.
What sign convention should you use for projectiles?
A projectile problem needs a consistent sign convention chosen before starting. The most reliable is upward as positive, so a = −9.8 m s⁻² throughout and any displacement under the launch point comes out negative automatically; for a horizontal launch, taking downward as positive can keep the arithmetic clean.
Why is time the link between horizontal and vertical motion?
Time is the one quantity shared by the horizontal and vertical motion of a projectile, because the object spends the same time in the air whichever direction is considered. Time is found from the vertical suvat equations first, then substituted into the horizontal equation to find the range.
Why is the path of a projectile a parabola?
A projectile's path is a parabola because its horizontal displacement grows linearly with time while its vertical displacement grows with t². Combining the two gives the characteristic curved path, which has symmetry when the launch and landing heights are equal.
How do you find the time to reach maximum height?
The time to reach maximum height comes from v = u + at with the vertical velocity v = 0 at the top. For a vertical launch speed of 10.0 m s⁻¹, 0 = 10.0 − 9.8t, so t = 1.02 s, which is half the total flight time when launch and landing heights are equal.
How do you find the range of a projectile launched at an angle?
For a projectile launched at an angle, the velocity is resolved first. A ball launched at 20 m s⁻¹ at 30° to the horizontal has components of 17.3 m s⁻¹ horizontally and 10.0 m s⁻¹ vertically; the vertical motion gives a flight time of 2.04 s, so the range is 17.3 × 2.04 = 35.3 m.
What is the initial vertical velocity of a horizontally launched projectile?
A projectile launched horizontally has zero initial vertical velocity, so in s = ut + ½at² the ut term vanishes and the vertical drop is ½gt². Falling 45 m therefore takes t² = 45 ÷ 4.9, giving t = 3.03 s for any horizontal speed.