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 $U$ at an angle $\alpha$ above the horizontal can be resolved into components.

Maths; Projectiles; KS5 Year 13; Projectiles: Horizontal and vertical components

Horizontal component:

U \cos \alpha

Vertical component:

U \sin \alpha

Example 1

A ball is thrown at $8\ m.s^{-1}$  at $60 \degree$ above the horizontal.

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

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

i) For the horizontal and vertical components:

$U=8\ m.s^{-1}$ and $\alpha=60\degree$ 

$u_x=8\cos{60=}\ \underline{4\ m.s^{-1}}$ 

$u_y=8\sin{60}=\underline{4\sqrt{3}\ m.s^{-1}}$ 

ii) For the horizontal and vertical displacement:

\begin{aligned}s_x&=vt\\s_x&=4\times 1\\ s_x&=\underline{4\ m} \end{aligned}

\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}

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 $\bold i, \bold j$ 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 $U = (2 \bold i + 7 \bold j) \ m.s^{-1}$ . 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:

$|U| = \sqrt{2^2+7^2} = \underline{\sqrt{53} \ m.s^{-1}}$ 

Use trigonometry to work out the angle of projection:

$\begin{aligned}\tan (\alpha) &= \dfrac {7} {2} \\\\\alpha &= \tan^{-1} \Big(\dfrac {7}{2}\Big) \\\\\alpha &=\underline{74.1 \degree}\end{aligned}$