Projectiles

Projectiles

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Projectiles

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

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

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

Horizontal:

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

Vertical:

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

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

gives

\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}

gives

\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

Example Questions

\underline{u} = 5 gives

Horizontally:

5\cos 60° = 2.5\text{ ms}^{-1}

Vertically:

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

Using

v = u + at

we have

0 = 14.7 - 9.8t

giving

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})

and

\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})

Meaning

125 - 30t + t^2 = 0

and

24.5t - 4.9t^2 = 0

This gives t = 5\text{ seconds}.

Related Topics

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SUVAT Equations

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Additional Resources

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Exam Tips Cheat Sheet

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Formula Booklet

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