A LevelAQAEdexcelOCR

Projectiles Revision


We model projectile motion in two components, horizontal and vertical.

Make sure you are happy with the following topics before continuing.

A LevelAQAEdexcelOCR

Splitting Velocity into Components

Using trigonometry, we convert a standard projectile motion into its two components.

Generally, we have a particle fired with a velocity u at an angle of \textcolor{orange}{\alpha}, which gives


u\cos \textcolor{orange}{\alpha}


u\sin \textcolor{orange}{\alpha}

From here, we can use either method of modelling motion – SUVAT or integration/differentiation. We should use these piecewise, meaning, our equations in the vertical component are not the same equations in the horizontal component.

A LevelAQAEdexcelOCR

Finding a Maximum Height and Maximum Velocity

Remember, we can also find a maximum or minimum displacement by differentiating and finding the time \textcolor{purple}{t} where the velocity of our object is 0.

We can also find a maximum or minimum velocity by differentiating again and finding a time \textcolor{purple}{t} where the acceleration, \textcolor{blue}{a} = 0.

A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

Example: Projectiles in Vector Notation

We can also use vectors to make projectile motion much neater.

So, for example, say a ball is thrown off of a cliff with a velocity of (15\textbf{i} + 7\textbf{j})\text{ ms}^{-1} with \textbf{i} its horizontal velocity, and \textbf{j} its upward vertical velocity. Assume that the ball accelerates due to gravity and experiences no air resistance. Given it is in the air for \textcolor{purple}{t} = \textcolor{purple}{5}\text{ seconds}, how tall is the cliff, what horizontal distance does the ball travel and what is its final velocity?

Assume g = 10\text{ ms}^{-2}.

[4 marks]

\textcolor{limegreen}{\underline{s}} = \underline{u}\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{\underline{a}}\textcolor{purple}{t}^2


\textcolor{limegreen}{\underline{s}} = \textcolor{purple}{5}(15\textbf{i} + 7\textbf{j}) + \dfrac{\textcolor{purple}{25}}{2}(\textcolor{blue}{-10\textbf{j}}) = \textcolor{limegreen}{75\textbf{i} - 90\textbf{j}}

So, the ball travels \textcolor{limegreen}{75}\text{ m} horizontally, and the cliff is \textcolor{limegreen}{90}\text{ m} tall.

\textcolor{red}{\underline{v}} = \underline{u} + \textcolor{blue}{\underline{a}}\textcolor{purple}{t}


\textcolor{red}{\underline{v}} = (15\textbf{i} + 7\textbf{j}) - (\textcolor{blue}{10} \times \textcolor{purple}{5})\textbf{j} = \textcolor{red}{15\textbf{i} - 43\textbf{j}}\text{ ms}^{-1}

A LevelAQAEdexcelOCR

Projectiles Example Questions

\underline{u} = 5 gives


5\cos 60° = 2.5\text{ ms}^{-1}


5\sin 60° = 4.33\text{  ms}^{-1}\text{ (to }2\text{ dp)}


v = u + at

we have

0 = 14.7 - 9.8t


t = 1.5\text{ seconds}

Substituting this into

s = ut + \dfrac{1}{2}at^2

we can prove that

\begin{aligned}s&=(14.7 \times 1.5) + \left( \dfrac{1}{2} \times -9.8 \times 1.5^2\right)\\[1.2em]&=11.025\text{ m}\end{aligned}

which is greater than 11\text{ m}, as required.

\underline{u} = (30\textbf{i} + 24.5\textbf{j})


\underline{a} = (-2\textbf{i} - 9.8\textbf{j})\text{ ms}^{-2}


Using \underline{s} = \underline{u}t + \dfrac{1}{2}\underline{a}t^2 gives

125\textbf{i} = (30t\textbf{i} + 24.5t\textbf{j}) + (-t^2\textbf{i} - 4.9t^2\textbf{j})


125 - 30t + t^2 = 0


24.5t - 4.9t^2 = 0

This gives t = 5\text{ seconds}.

Additional Resources


Exam Tips Cheat Sheet

A Level

Formula Booklet

A Level

You May Also Like...

MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.

View Product

Related Topics


SUVAT Equations

A Level