Projectiles

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Projectiles

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

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

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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 uu at an angle of αtextcolor{orange}{alpha}, which gives

Horizontal:

ucosαucos textcolor{orange}{alpha}

Vertical:

usinαusin 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.

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Finding a Maximum Height and Maximum Velocity

Remember, we can also find a maximum or minimum displacement by differentiating and finding the time ttextcolor{purple}{t} where the velocity of our object is 00.

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

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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 (15i+7j) ms1(15textbf{i} + 7textbf{j})text{ ms}^{-1} with itextbf{i} its horizontal velocity, and jtextbf{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 t=5 secondstextcolor{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 ms2g = 10text{ ms}^{-2}.

[4 marks]

s=ut+12at2textcolor{limegreen}{underline{s}} = underline{u}textcolor{purple}{t} + dfrac{1}{2}textcolor{blue}{underline{a}}textcolor{purple}{t}^2

gives

s=5(15i+7j)+252(10j)=75i90jtextcolor{limegreen}{underline{s}} = textcolor{purple}{5}(15textbf{i} + 7textbf{j}) + dfrac{textcolor{purple}{25}}{2}(textcolor{blue}{-10textbf{j}}) = textcolor{limegreen}{75textbf{i} – 90textbf{j}}

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

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

gives

v=(15i+7j)(10×5)j=15i43j ms1textcolor{red}{underline{v}} = (15textbf{i} + 7textbf{j}) – (textcolor{blue}{10} times textcolor{purple}{5})textbf{j} = textcolor{red}{15textbf{i} – 43textbf{j}}text{ ms}^{-1}

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Projectiles Example Questions

Question 1: A particle is fired at a velocity of 5 ms15text{ ms}^{-1} at an angle of 60°60°. What are the horizontal and vertical components of this velocity?

[2 marks]

A Level AQAEdexcelOCR

u=5underline{u} = 5 gives

Horizontally:

5cos60°=2.5 ms15cos 60° = 2.5text{ ms}^{-1}

Vertically:

5sin60°=4.33  ms1 (to 2 dp)5sin 60° = 4.33text{  ms}^{-1}text{ (to }2text{ dp)}

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Question 2: A football is kicked directly upwards with a velocity of 14.7 ms114.7text{ ms}^{-1}. Show that the ball’s height exceeds 11 m11text{ m}, and that this maximum height occurs when t=1.5 secondst = 1.5text{ seconds}.

 

Use the value g=9.8 ms1g=9.8 text{ ms}^{-1}

[2 marks]

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Using

v=u+atv = u + at

we have

0=14.79.8t0 = 14.7 – 9.8t

giving

t=1.5 secondst = 1.5text{ seconds}

Substituting this into

s=ut+12at2s = ut + dfrac{1}{2}at^2

we can prove that

s=(14.7×1.5)+(12×9.8×1.52)=11.025 mbegin{aligned}s&=(14.7 times 1.5) + left( dfrac{1}{2} times -9.8 times 1.5^2right)[1.2em]&=11.025text{ m}end{aligned}

which is greater than 11 m11text{ m}, as required.

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Question 3: A golf ball is hit with an initial velocity of (30i+24.5j) ms1(30textbf{i} + 24.5textbf{j})text{ ms}^{-1}, where itextbf{i} represents the forward direction, and jtextbf{j} represents upward vertical motion. Given that there is a constant headwind, impacting the ball’s acceleration by 2 ms2-2text{ ms}^{-2}, and the ball lands 125 m125text{ m} from the tee, how long is it in flight for?

[5 marks]

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

and

a=(2i9.8j) ms2underline{a} = (-2textbf{i} – 9.8textbf{j})text{ ms}^{-2}

 

Using s=ut+12at2underline{s} = underline{u}t + dfrac{1}{2}underline{a}t^2 gives

125i=(30ti+24.5tj)+(t2i4.9t2j)125textbf{i} = (30ttextbf{i} + 24.5ttextbf{j}) + (-t^2textbf{i} – 4.9t^2textbf{j})

Meaning

12530t+t2=0125 – 30t + t^2 = 0

and

24.5t4.9t2=024.5t – 4.9t^2 = 0

This gives t=5 secondst = 5text{ seconds}.

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Additional Resources

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Exam Tips Cheat Sheet

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Formula Booklet

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Specification Points Covered

Q5 – Model motion under gravity in a vertical plane using vectors; projectiles

Related Topics

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SUVAT Equations

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