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Projectiles: Horizontal and vertical components

Projectiles: Horizontal and vertical components

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Summary

Projectiles: Horizontal and vertical components

In a nutshell

Rather than projecting a particle just horizontally or vertically, it is possible to project a particle at any angle. The velocity of the projected particle can be resolved into horizontal and vertical components, according to the angle of projection. The horizontal and vertical components can then be used in further calculations in a projectile motion question.



Horizontal and vertical components

A particle projected at a constant velocity UUU at an angle α\alphaα above the horizontal can be resolved into components. 


Maths; Projectiles; KS5 Year 13; Projectiles: Horizontal and vertical components
Horizontal component: Ucos⁡αU \cos \alphaUcosα​​
Vertical component: Usin⁡αU \sin \alphaUsinα​​


Example 1

A ball is thrown at 8 m.s−18\ m.s^{-1}8 m.s−1​ at 60°60 \degree60° above the horizontal. 

i) Find the horizontal and vertical components of the initial velocity. 

ii) Find the horizontal and vertical displacement at t=1 st=1\ st=1 s.


i) For the horizontal and vertical components:

U=8 m.s−1U=8\ m.s^{-1}U=8 m.s−1​ and α=60°\alpha=60\degreeα=60°​


ux=8cos⁡60= 4 m.s−1‾u_x=8\cos{60=}\ \underline{4\ m.s^{-1}}ux​=8cos60= 4 m.s−1​​​

uy=8sin⁡60=43 m.s−1‾u_y=8\sin{60}=\underline{4\sqrt{3}\ m.s^{-1}}uy​=8sin60=43​ m.s−1​​​


ii) For the horizontal and vertical displacement:

​sx=vtsx=4×1sx=4 m‾\begin{aligned}s_x&=vt\\s_x&=4\times 1\\ s_x&=\underline{4\ m} \end{aligned}sx​sx​sx​​=vt=4×1=4 m​​​​
​sy=ut+12at2sy=43×1+12×(−9.8)×12sy=2.03 m‾ (3 s.f)\begin{aligned}s_y&=ut+\dfrac 1 2 at^2\\s_y&=4\sqrt{3}\times 1 + \dfrac 1 2 \times (-9.8)\times1^2\\s_y&=\underline{2.03\ m} \ (3\thinspace s.f)\end{aligned}sy​sy​sy​​=ut+21​at2=43​×1+21​×(−9.8)×12=2.03 m​ (3s.f)​​​



Velocity given in component form

In some cases, the horizontal and vertical components of the initial velocity will be given in the question. When the components are given, the velocity can be written as a vector or using i,j\bold i, \bold ji,j notation. It is possible to work out the magnitude of the velocity and the angle of projection using Pythagoras' theorem and trigonometry.


Example 2

A particle is projected with velocity U=(2i+7j) m.s−1U = (2 \bold i + 7 \bold j) \ m.s^{-1}U=(2i+7j) m.s−1. Calculate the magnitude of the initial velocity and the angle of projection.


The magnitude of the initial velocity can be calculated by applying Pythagoras' theorem:

∣U∣=22+72=53 m.s−1‾|U| = \sqrt{2^2+7^2} = \underline{\sqrt{53} \ m.s^{-1}}∣U∣=22+72​=53​ m.s−1​​​


Use trigonometry to work out the angle of projection:

​tan⁡(α)=72α=tan⁡−1(72)α=74.1°‾\begin{aligned}\tan (\alpha) &= \dfrac {7} {2} \\\\\alpha &= \tan^{-1} \Big(\dfrac {7}{2}\Big) \\\\\alpha &=\underline{74.1 \degree}\end{aligned}tan(α)αα​=27​=tan−1(27​)=74.1°​​​​


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Projectiles: Horizontal and vertical components

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FAQs - Frequently Asked Questions

If the initial velocity of a projected particle is given in vector form, how do you find the angle of projection?

If the initial velocity is given as U = (xi + yj) m/s, then the angle of projection can be found by using tan^(-1) (y/x).

What do you do when a particle is projected at an angle?

The velocity of the projected particle can be resolved into horizontal and vertical components, according to the angle of projection.

How do you resolve the velocity of a particle when it's projected at an angle?

When a particle is projected with initial velocity U, at an angle x above the horziontal. The horizontal component is Ucos(x) and the vertical component is Usin(x).

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