Calculating Projectile Motion Revision Notes | A Level AQA Physics

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 $g = 9.8 \text{ m s}^{-2}$.

1. 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.
2. $s = ut + \frac{1}{2}at^2$ $45 = (0)t + \frac{1}{2}(9.8)t^2$ $45 = 4.9t^2$ — 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.
3. $t^2 = \frac{45}{4.9} = 9.1837...$ $t = \sqrt{9.1837} = 3.03 \text{ s}$ (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]
4. $s_x = u_x \times t = 15 \times 3.03 = 45.5 \text{ m}$ (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]
5. 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.

Answer: (a) $t = 3.03 \text{ s}$ (b) Range $= 45.5 \text{ m}$

5. Worked Example: Angled Launch

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 $g = 9.8 \text{ m s}^{-2}$.

1. $u_x = 20\cos 30° = 20 \times 0.866 = 17.3 \text{ m s}^{-1}$ $u_y = 20\sin 30° = 20 \times 0.500 = 10.0 \text{ m s}^{-1}$ (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]
2. $v^2 = u^2 + 2as$ 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]
3. $s = u_y t + \frac{1}{2}at^2$ $0 = 10.0t + \frac{1}{2}(-9.8)t^2$ $0 = 10.0t - 4.9t^2$ $0 = t(10.0 - 4.9t)$ $t = 0$ (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]
4. $s_x = u_x \times t = 17.3 \times 2.04 = 35.3 \text{ m}$ (3 s.f.) — Horizontal acceleration is zero, so $s_x = u_x t$. Substituting $17.3 \times 2.04$: $17.3 \times 2 = 34.6$ and $17.3 \times 0.04 = 0.692$, giving $34.6 + 0.692 = 35.3 \text{ m}$. [A1]
5. 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.