Projectiles

This lesson covers projectile motion at A-Level, where objects move in two dimensions under the influence of gravity alone. The key principle is that horizontal and vertical components of motion are treated independently.

Assumptions in the Projectile Model

The standard projectile model assumes:

The object is a particle (no air resistance, no spin).
The only force acting is gravity (vertically downwards, $g = 9.8$ m/s $^2$ ).
Horizontal velocity is constant (no horizontal forces).
Vertical motion has constant acceleration $g$ downwards.

Exam Tip: If asked to "state assumptions", always mention: the object is modelled as a particle, air resistance is neglected, and $g$ is constant. You should also note that the ground is assumed to be flat and horizontal.

Resolving Initial Velocity

If an object is projected at speed $U$ at angle $\theta$ above the horizontal:

Component	Value
Horizontal	$U_x = U\cos\theta$
Vertical	$U_y = U\sin\theta$

Equations of Motion

Horizontal Motion (constant velocity)

$x = U\cos\theta \cdot t$

There is no horizontal acceleration, so horizontal velocity remains $U\cos\theta$ throughout.

Vertical Motion (constant acceleration $-g$ )

Using SUVAT with $u = U\sin\theta$ , $a = -g$ :

$v_y = U\sin\theta - gt$ $y = U\sin\theta \cdot t - \frac{1}{2}gt^2$ $v_y^2 = (U\sin\theta)^2 - 2gy$

Key Results

Time of Flight

The total time in the air (landing at the same height as launch):

$T = \frac{2U\sin\theta}{g}$

Maximum Height

At the highest point, $v_y = 0$ :

$H = \frac{U^2 \sin^2\theta}{2g}$

Horizontal Range

The horizontal distance travelled during the time of flight:

$R = \frac{U^2 \sin 2\theta}{g}$

The maximum range occurs when $\theta = 45\degree$ .

Exam Tip: These formulae are useful for quick checks but you should be able to derive them from first principles. In multi-part questions, the examiner often expects you to work through the SUVAT equations step by step rather than quote a result directly.

The Trajectory Equation

Eliminating $t$ between the horizontal and vertical equations gives the equation of the trajectory:

$y = x\tan\theta - \frac{gx^2}{2U^2\cos^2\theta}$

This is a quadratic in $x$ , confirming the path is a parabola.

Projectiles Launched Horizontally

If the object is projected horizontally from a height $h$ (so $\theta = 0$ ):

$U_x = U$ , $U_y = 0$
$x = Ut$
$y = -\frac{1}{2}gt^2$ (measuring y downwards from launch point)

Example: A ball is projected horizontally at 10 m/s from a cliff 45 m high. Find the time to hit the ground and the horizontal distance.

Vertical: $45 = \frac{1}{2}(9.8)t^2$ , so $t^2 = \frac{90}{9.8} = 9.18$ , $t = 3.03$ s.

Horizontal: $x = 10 \times 3.03 = 30.3$ m.

Velocity at Any Point

The velocity at time $t$ has components:

Horizontal: $v_x = U\cos\theta$ (constant)
Vertical: $v_y = U\sin\theta - gt$

The speed at time $t$ is: $|v| = \sqrt{v_x^2 + v_y^2}$

The angle below the horizontal is: $\alpha = \arctan\left(\frac{|v_y|}{v_x}\right)$

Summary

Projectile motion is analysed by treating horizontal and vertical components independently.
Horizontal: constant velocity (no acceleration).
Vertical: constant acceleration $g$ downwards.
Resolve initial velocity into components: $U\cos\theta$ and $U\sin\theta$ .
The path of a projectile is a parabola.
Time of flight, maximum height, and range can be found using SUVAT equations.

Exam Tip: Always set up a clear coordinate system (e.g., horizontal = positive x, vertical upwards = positive y). Write down horizontal and vertical equations separately. Many errors come from mixing components or sign errors with $g$ .

A-Level Deep Dive: Projectile Motion

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Section T (Year 2 content): Kinematics. This sub-strand covers projectile motion — modelling motion under gravity in a vertical plane using vectors (refer to the official AQA 7357 specification document for exact wording). Projectile motion is the synoptic capstone of Mechanics — it requires confident SUVAT application in two independent directions, vector resolution, trigonometric identities, and (for the trajectory equation) parametric-to-Cartesian conversion. It is examined in Paper 3 only, but draws on Pure content from section E (Trigonometry: $\sin 2\alpha = 2\sin\alpha\cos\alpha$ for the range formula), section J (Vectors: resolving $\mathbf{u}$ into components) and section H (Parametric equations: eliminating $t$ to obtain $y$ in terms of $x$ ). The AQA formula booklet does not list the range or trajectory formulae — they must be derived on demand from $s = ut + \tfrac{1}{2}at^2$ .

Worked example with full mark scheme

Question (8 marks):

A ball is projected from a point on horizontal ground with initial speed $U\,\text{m s}^{-1}$ at an angle $\alpha$ above the horizontal, where $U = 28$ and $\tan\alpha = \tfrac{3}{4}$ . Take $g = 9.8\,\text{m s}^{-2}$ and ignore air resistance.

(a) Find the time the ball takes to return to the ground. (3)

(b) Find the horizontal range. (2)

(d) Show that the trajectory satisfies $y = \tfrac{3}{4}x - \dfrac{49 x^2}{2 \cdot 28^2 \cdot (4/5)^2}$ and simplify. (1)

Solution with mark scheme:

Since $\tan\alpha = 3/4$ , we have $\sin\alpha = 3/5$ and $\cos\alpha = 4/5$ (3-4-5 triangle). Therefore:

Horizontal component: $u_x = U\cos\alpha = 28 \cdot \tfrac{4}{5} = 22.4\,\text{m s}^{-1}$ .
Vertical component: $u_y = U\sin\alpha = 28 \cdot \tfrac{3}{5} = 16.8\,\text{m s}^{-1}$ .

B1 (implicit) for resolving correctly into components — usually awarded inside the first M-mark.

(a) Step 1 — apply vertical SUVAT with $s = 0$ (returning to ground level).

Taking up as positive: $a = -g = -9.8$ , $u = u_y = 16.8$ , $s = 0$ .

$s = ut + \tfrac{1}{2}at^2 \implies 0 = 16.8\,t - 4.9\,t^2 = t(16.8 - 4.9\,t)$

M1 — correct vertical SUVAT setup with displacement zero. Common error: students set $v = 0$ instead, which gives time to maximum height (half the answer). That earns nothing on this part.

Step 2 — solve.

$t = 0$ (start) or $t = 16.8/4.9 = 24/7 \approx 3.43\,\text{s}$ .

A1 — correct non-trivial root. Reject $t = 0$ explicitly.

A1 — final answer $t = 24/7\,\text{s}$ or $3.43\,\text{s}$ (3 s.f.).

(b) Horizontal range = horizontal velocity × time of flight.

$R = u_x \cdot t = 22.4 \cdot \tfrac{24}{7} = \tfrac{537.6}{7} = 76.8\,\text{m}$

M1 — using $R = u_x \, T$ with the time from (a). The horizontal motion is uniform ( $a_x = 0$ ), so $s_x = u_x t$ — no acceleration term.

A1 — $R = 76.8\,\text{m}$ .

$0 = u_y^2 - 2g\,H \implies H = \dfrac{u_y^2}{2g} = \dfrac{16.8^2}{19.6} = \dfrac{282.24}{19.6} = 14.4\,\text{m}$

M1 — using $v^2 = u^2 + 2as$ with $v_y = 0$ at the apex.

A1 — $H = 14.4\,\text{m}$ .

(d) Trajectory equation. Eliminate $t$ between horizontal and vertical equations.

From $x = u_x t$ : $t = x / u_x = x / (U\cos\alpha)$ .

Substitute into $y = u_y t - \tfrac{1}{2} g t^2$ :

$y = U\sin\alpha \cdot \dfrac{x}{U\cos\alpha} - \tfrac{1}{2} g \cdot \dfrac{x^2}{U^2\cos^2\alpha} = x\tan\alpha - \dfrac{g\,x^2}{2U^2\cos^2\alpha}$

With $\tan\alpha = 3/4$ , $\cos\alpha = 4/5$ , $g = 9.8$ , $U = 28$ :

$y = \tfrac{3}{4}x - \dfrac{9.8\,x^2}{2 \cdot 784 \cdot 16/25} = \tfrac{3}{4}x - \dfrac{9.8 \cdot 25\,x^2}{2 \cdot 784 \cdot 16} = \tfrac{3}{4}x - \dfrac{245\,x^2}{25088}$

B1 — printed result obtained, simplification shown.

Total: 8 marks (M4 A4 B1 distributed as shown).

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): A stone is thrown from the top of a vertical cliff of height $45\,\text{m}$ with initial speed $20\,\text{m s}^{-1}$ at an angle of $30°$ above the horizontal. The stone lands on horizontal ground at the foot of the cliff. Take $g = 9.8\,\text{m s}^{-2}$ and model the stone as a particle moving freely under gravity.

(a) Find the time taken for the stone to reach the ground. (4)

(b) Find the horizontal distance from the foot of the cliff to the point where the stone lands. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO3.1b) — modelling: identify vertical SUVAT with $s = -45$ (taking up as positive, ground is below launch), $u = 20\sin 30° = 10$ , $a = -9.8$ .
M1 (AO1.1b) — substitute into $s = ut + \tfrac{1}{2}at^2$ : $-45 = 10\,t - 4.9\,t^2$ , rearrange to $4.9\,t^2 - 10\,t - 45 = 0$ .
M1 (AO1.1b) — apply quadratic formula: $t = \dfrac{10 \pm \sqrt{100 + 882}}{9.8} = \dfrac{10 \pm \sqrt{982}}{9.8}$ .
A1 (AO1.1b) — take positive root: $t = (10 + 31.34)/9.8 \approx 4.22\,\text{s}$ .

(b)

M1 (AO1.1a) — horizontal distance $= 20\cos 30° \cdot t = 10\sqrt{3} \cdot 4.22$ .
A1 (AO1.1b) — $\approx 73.1\,\text{m}$ (3 s.f.).

Total: 6 marks split AO1 = 5, AO3 = 1. AQA Paper 3 projectile questions allocate the AO3 (modelling) mark for choosing the correct sign convention and origin — once that decision is made, the rest is procedural AO1.

Synoptic links

Connects to:

Section E — Trigonometry (double-angle identity): the range formula $R = \dfrac{U^2 \sin 2\alpha}{g}$ is derived using $2\sin\alpha\cos\alpha = \sin 2\alpha$ . Without confident manipulation of the double-angle identity, the elegant single-formula result is invisible — students must instead compute time of flight and multiply by horizontal speed every time.
Section J — Vectors: initial velocity $\mathbf{u} = U\cos\alpha\,\mathbf{i} + U\sin\alpha\,\mathbf{j}$ . Acceleration is $\mathbf{a} = -g\,\mathbf{j}$ . Position vector at time $t$ : $\mathbf{r}(t) = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$ . The vector formulation makes the independence of horizontal and vertical motion structurally obvious — they are different components of the same vector equation.
Section H — Parametric equations: the trajectory $y(x)$ is obtained by eliminating the parameter $t$ from the parametric pair $(x(t), y(t))$ . This is the same technique used to convert $x = \cos\theta$ , $y = \sin\theta$ to $x^2 + y^2 = 1$ in Pure section H — projectile motion is parametric curves applied.
Section R — Kinematics (Year 1): SUVAT $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ are applied separately to the horizontal direction (with $a = 0$ , so $s = u_x t$ ) and the vertical direction (with $a = -g$ ). The 2D problem is two 1D problems run in parallel.
Section O — Forces and Newton's laws: the only force is gravity (assuming no air resistance), so $F = mg$ acts downward and $\mathbf{a} = -g\,\mathbf{j}$ regardless of mass. The mass-independence of trajectories is a synoptic Newton-II observation that surprises students until they see it derived.
Modelling air resistance (Year 2 extension): introducing a drag force $-k\mathbf{v}$ couples the horizontal and vertical equations, and the trajectory ceases to be parabolic. AQA Mech section T explicitly mentions modelling assumptions — air resistance is the modelling assumption to critique.

Mark-scheme literacy

Projectile questions on AQA Paper 3 split AO marks more evenly than pure topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Resolving velocity, applying SUVAT vertically and horizontally, computing time of flight / range / max height
AO2 (reasoning / interpretation)	15–20%	Justifying $v_y = 0$ at maximum height, explaining why horizontal motion is uniform, deriving the trajectory equation
AO3 (problem-solving / modelling)	25–30%	Choosing sign conventions, setting up axes, critiquing the no-air-resistance assumption, modelling cliff/inclined launches

Examiner-rewarded phrasing: "taking up as positive, with the launch point as the origin"; "since $v_y = 0$ at the maximum height"; "the horizontal acceleration is zero, so horizontal velocity is constant at $U\cos\alpha$ "; "modelling the projectile as a particle". Phrases that lose marks: forgetting to specify the sign convention before substituting; applying $v = 0$ at the landing point rather than the apex; treating the trajectory as a straight line because "no horizontal force" gets misread as "no horizontal motion".

A specific AQA pattern: questions phrased "the projectile is modelled as a particle moving freely under gravity" encode three modelling assumptions — point mass, no air resistance, uniform gravity — and Paper 3 questions sometimes ask candidates to critique these assumptions for an extra AO3 mark. Always remember: "modelling air resistance as zero overstates the range, particularly for low-density projectiles or high speeds".

Grade-band model answers

3-mark question

Question: A particle is projected horizontally from a height of $1.8\,\text{m}$ with speed $5\,\text{m s}^{-1}$ . Find the horizontal distance travelled before it hits the ground. Take $g = 9.8\,\text{m s}^{-2}$ .

Grade C response (~180 words):

Vertically: $s = -1.8$ , $u = 0$ , $a = -9.8$ . Using $s = ut + \tfrac{1}{2}at^2$ : $-1.8 = -4.9\,t^2$ , so $t^2 = 1.8/4.9 = 0.367$ , giving $t = 0.606\,\text{s}$ .

Horizontally: distance $= 5 \cdot 0.606 = 3.03\,\text{m}$ .

Examiner commentary: Full marks (3/3). Sign convention is consistent (down negative, displacement $-1.8$ , acceleration $-9.8$ ), giving a positive $t^2$ . The candidate correctly identifies that horizontal velocity is constant at $5\,\text{m s}^{-1}$ because $a_x = 0$ , then multiplies by time of flight. Working is brief but each step is justified. This is the standard Grade C answer for a procedural projectile question — efficient and correct. Note that a careless candidate often writes $s = +1.8$ (treating the magnitude of fall) while keeping $a = -9.8$ , producing a negative $t^2$ and panic. The discipline here is choosing one direction as positive before substituting any number.

Grade A response (~230 words):*

Take the launch point as the origin and upward as positive. Initial velocity components: $u_x = 5\,\text{m s}^{-1}$ , $u_y = 0$ .

Projectiles

Projectiles

Assumptions in the Projectile Model

Resolving Initial Velocity

Equations of Motion

Horizontal Motion (constant velocity)

Vertical Motion (constant acceleration −g-g−g)

Key Results

Time of Flight

Maximum Height

Horizontal Range

The Trajectory Equation

Projectiles Launched Horizontally

Velocity at Any Point

Summary

A-Level Deep Dive: Projectile Motion

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

More in Mathematics

Vertical Motion (constant acceleration $-g$ )