Mastering Projectile Motion

Projectile motion is the point in Mechanics where the diagrams start to feel like real life. Kicks, throws, serves, long passes, water jets, fireworks — they all follow the same simple idea: motion in two perpendicular directions at once. Horizontal and vertical. Independent, but sharing the same time.

Examiners on AQA, Edexcel, and OCR will test whether you can split the motion into those two parts, keep your signs straight, and choose the right kinematics formula at the right moment. If you can do that calmly, projectiles turn into very manageable marks. Don’t worry if the first few look wordy — once you settle into a routine, the pattern shows up fast.

🔙 Previous topic:

“Go back to basic projectile motion first.”

🧠 Key idea in one sentence

A projectile moves with constant horizontal velocity (no horizontal forces in the basic model) and constant vertical acceleration (gravity, acting downwards). The two components are independent, but the time t is the same for both.

That’s the whole game. Everything else is bookkeeping.

🔩 Core formulas (plain text, copy straight into your notes)

Let the launch speed be $u$ and the launch angle above the horizontal be $θ\theta$ .

Decompose the initial velocity:

Horizontal component:
$u_x = u \cos\theta$
Vertical component:
$u_y = u \sin\theta$

Horizontal motion (no acceleration in the basic model):

$x = u_x t$

Vertical motion (constant acceleration $a = - g$ ):

$v_y = u_y - g t$

$y = u_y t – \tfrac12 g t^2$

$v_y^2 = u_y^2 - 2 g y$

Useful results when launch and landing heights are the same ( $y = 0$ ):

Time of flight:
$T = \frac{2u_y}{g}$
Range:
$R = \frac{u^2 \sin(2\theta)}{g}$
Maximum height:
$H = \frac{u_y^2}{2g}$

Teacher aside:
Write $9.8\ \text{m/s}^2$ unless the paper specifies $9.81$ or $10$ . Use the value given in the exam and stay consistent throughout.

🧩 How to think about any projectile (step-by-step method)

Sketch and label. A quick arc, axes, launch point, angle, and what they’re actually asking for. Mark “up is positive” to keep your signs tidy.
Split the initial velocity. Write u_x and u_y clearly before touching equations.
Pick the right component for the question.
- Want range or time in the air? Usually start with vertical (y equation) to get t, then use x = u_x t.
- Want maximum height? Use vertical only: set v_y = 0 and use v_y^2 = u_y^2 − 2 g y.
- Want launch angle for a given range? Combine R = u^2 sin(2 theta) / g with any height info provided.
Keep t consistent. One time variable; horizontal and vertical share it.
Answer the exact wording. If it says “speed on impact,” combine components at the end: speed = sqrt(v_x^2 + v_y^2) and direction = arctan(v_y / v_x).

Sense-check. If your time is negative or your range is bigger than it could plausibly be, pause and recheck signs.

🧪 Worked Example 1 – classic ground-to-ground shot

A ball is launched at speed $m/s20\,\text{m/s}$ at an angle $35∘35^\circ$ above the horizontal. Ignore air resistance.

Decompose the initial velocity:

$u_x = 20\cos 35^\circ \approx 16.38\ \text{m/s}$

$u_y = 20\sin 35^\circ \approx 11.48\ \text{m/s}$

(a) Time of flight (ground to ground, so $y = 0$ at landing):

$T = \frac{2u_y}{g} = \frac{2 \times 11.48}{9.8} \approx 2.34\ \text{s}$

(b) Range:

$R = u_x T \approx 16.38 \times 2.34 \approx 38.3\ \text{m}$

(c) Maximum height (vertical only; set $v_y = 0$ at the top):

Using
$v_y^2 = u_y^2 - 2gH$

Set $v_y = 0$ :

$0 = 11.48^2 – 2 \times 9.8 \times H$

Thus,

$H = \frac{11.48^2}{2 \times 9.8} \approx 6.73\ \text{m}$

Teacher aside: Don’t round too early. Keep 3 significant figures in your working and round only at the end of each part.

🧪 Worked Example 2 – launched from a platform

A javelin is released from a height of $m1.6\,\text{m}$ with speed $m/s22\,\text{m/s}$ at $28∘28^\circ$ above the horizontal.

Components:

$u_x = 22\cos 28^\circ \approx 19.40\ \text{m/s}$

$u_y = 22\sin 28^\circ \approx 10.31\ \text{m/s}$

Vertical motion with initial height $my_0 = 1.6\,\text{m}$ ; ground is $y = 0$ .

Equation:

$y = y_0 + u_y t – \tfrac12 g t^2 = 0$

So:

$0 = 1.6 + 10.31 t - 4.9 t^2$

Rearrange:

$4.9 t^2 - 10.31 t - 1.6 = 0$

Quadratic discriminant:

$D = (10.31)^2 + 4(4.9)(1.6) \approx 137.7$

$\sqrt{D} \approx 11.74$

Time:

$t = \frac{10.31 \pm 11.74}{9.8}$

Positive root only:

$t \approx \frac{10.31 + 11.74}{9.8} \approx 2.25\ \text{s}$

(a) Time until it hits the ground

$t \approx 2.25\ \text{s}$

(b) Horizontal distance (range)

$R = u_x t \approx 19.40 \times 2.25 \approx 43.7\ \text{m}$

(c) Speed on impact

Horizontal velocity remains:

$v_x = 19.40\ \text{m/s}$

Vertical velocity at impact:

$v_y = u_y – gt \approx 10.31 – 9.8(2.25) \approx -11.74\ \text{m/s}$

Impact speed:

$v = \sqrt{v_x^2 + v_y^2} \approx \sqrt{19.40^2 + (-11.74)^2} \approx 22.7\ \text{m/s}$

Impact angle below horizontal:

$\theta = \arctan!\left(\frac{|v_y|}{v_x}\right) \approx \arctan!\left(\frac{11.74}{19.40}\right) \approx 31^\circ$

Sanity check: Impact speed is close to launch speed (air resistance ignored), which is physically sensible given the small release height.

🧩 Common exam twists (and how to handle them)

Different landing heights (we just did one). Use a vertical quadratic for $t$ ; the “range formula”
$\frac{u^2 \sin(2\theta)}{g}$
no longer applies directly.

Find angle for a given range. Use
$R = \frac{u^2 \sin(2\theta)}{g}$
only if launch and landing heights match. Remember that $sin⁡(2θ)\sin(2\theta)$ gives two possible angles in $⁣180∘0^\circ\!-\!180^\circ$ ; if both are physically plausible, the question’s context will decide.

Two-stage flight (hit a wall or clear a bar). Do vertical first to find the height at a particular $x$ , or solve for $t$ to reach the obstacle’s $x$ , then check $y$ at that $t$ .

Given the impact speed/direction, find launch conditions. Work backwards: use the final velocity components to find the time (vertical first), then recover $u_x$ and $u_y$ , and finally $u$ and $θ\theta$ .

Projectile from a moving platform. Add the platform’s velocity vector to the initial velocity before resolving into components.

Teacher aside: If you’re not sure where to start, write the two core equations:
$x = u_x t$
and
$y = u_y t – \tfrac12 g t^2$
(plus any initial height). Nine times out of ten the path becomes clear from there.

📈 Graphs: quick wins if you don’t rush

$\text{A } y\text{–}t \text{ graph (vertical position vs time) is a parabola opening downwards. The maximum occurs at } t = \frac{u_y}{g}.$

$\text{A } v_y\text{–}t \text{ graph is a straight line with gradient } -g \text{ and intercept } u_y. \text{ The signed area under the graph gives the vertical displacement.}$

$\text{An } x\text{–}t \text{ graph is a straight line with gradient } u_x \text{ since horizontal velocity is constant.}$

If the question provides a height–time or velocity–time sketch, grab the easy marks by reading off key times, intercepts, and areas. Label them on the graph—this earns method credit.

🧭 A reliable problem-solving checklist

$\text{Define positive directions (usually right and up).}$

$\text{Draw a small diagram and write } u_x \text{ and } u_y.$

$\text{Write the two component equations before substituting numbers.}$

$\text{Use the vertical equation to find } t \text{ if the range is unknown.}$

$x = u_x t \quad \text{(use this for horizontal distance once you have } t\text{).}$

$\text{For maximum height, use vertical motion only and set } v_y = 0.$

$\text{For impact speed/direction, combine horizontal and vertical components at the end.}$

$\text{Units: metres, seconds, m/s, m/s}^2\text{. Keep them consistent.}$

$\text{Round once per part; } 3 \text{ s.f. is standard unless stated otherwise.}$

⚠️ Common mistakes (and the easy fixes)

$\text{Forgetting to split } u. \text{ Do not use } u \text{ directly in the } x \text{ or } y \text{ equations; always use } u_x \text{ and } u_y.$

$\text{Mixing signs. Gravity is downward, so } a_y = -g. \text{ If up is positive, } v_y \text{ decreases over time.}$

$\text{Using the range formula when heights differ. It is only valid for level-to-level flight. Otherwise, use the vertical quadratic.}$

$\text{Two different } t\text{'s? There is only one time variable — shared by both } x \text{ and } y \text{ equations.}$

$\text{Rounding too early. Keep 3–4 significant figures in the working; round final answers sensibly.}$

🧰 Quick practice prompts (try these aloud)

u = 18 m/s at 40 degrees from ground level. Find range, flight time, and maximum height.
u = 25 m/s at 30 degrees from a 2.0 m platform. Time to ground and speed on impact.
A ball is kicked so that it passes over a 2.4 m crossbar 18 m away. Minimum launch speed if theta = 35 degrees.
A shot lands 45 m away after 2.1 s. Find u and theta. (Hint: use x first, then vertical to solve for theta.)

If you can do those in five to eight minutes each, you’re exam-ready.

📚 Revision plan (keeps the topic warm)

Five diagrams a day for a week. Sketch, label u_x/u_y, write the two equations, stop. Building that reflex is half the battle.
Board-specific past paper sets. Do three projectile questions from your exam board back-to-back; patterns appear quickly.
Explain to a friend. If you can state why time is shared but acceleration is different in each axis, you own the idea.

Mistake log. One line per error: “Used u instead of u_x” → fix: “split first.” It’s dull but it works.

💬 Mini-FAQ

Do I need the range formula?
It’s handy but optional. You can always get range by finding time from the vertical equation and then x = u_x t.

Which angle gives the maximum range (level ground)?
Theta = 45 degrees, because sin(2 theta) is maximal at 90 degrees. If release/landing heights differ, the optimum shifts — don’t blindly use 45.

How do I report the direction of impact?
Give a bearing like “31 degrees below the horizontal” or “angle of depression 31 degrees,” and include the speed from the vector magnitude.

🏁 Final thoughts and CTA

Projectile motion looks flashy, but it’s built from two straight-line kinematics problems running side by side. Split the velocity, keep time consistent, and let the vertical equation do the heavy lifting. After a handful of full solutions, you’ll start predicting the shape of the answer before you even write it down — which is exactly where you want to be before the exam.

Start your revision for A Level Maths today with our A Level Maths half-term revision course— we cover projectiles for AQA, Edexcel, and OCR step by step, with board-style questions, worked solutions, and quick checks that build speed and accuracy.

About the Author

S. Mahandru is the Head of Mathematics at Exam.tips, specialising in A Level and GCSE Mathematics education. With over a decade of classroom and online teaching experience, he has helped thousands of students achieve top results through clear explanations, practical examples, and applied learning strategies.

Updated: November 2025

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“Now, strengthen your force-resolution skills.”

Mastering Projectile Motion