Mathematics/Lessons/Projectiles/Horizontal and Vertical Components
Mechanics · Projectiles · Year 2

Horizontal and Vertical Components

When a particle is launched at an angle rather than straight sideways, its initial velocity has both a horizontal and a vertical part. Resolving into these two components is what lets you apply everything from the last lesson.

9 min read 2 worked examples

Resolving velocity into components

When a particle is projected with speed $U$ at an angle $\alpha$ above the horizontal, it's useful to think of that single velocity as two separate velocities happening at the same time: one horizontal, one vertical. This is exactly the same idea as resolving a force, applied to a velocity vector instead.

Drawing $U$ as the hypotenuse of a right-angled triangle, with $\alpha$ measured from the horizontal:

Resolving the initial velocity

Horizontal component: $\;u_x = U\cos\alpha$

Vertical component: $\;u_y = U\sin\alpha$

Once you have $u_x$ and $u_y$, the particle behaves exactly as in horizontal projection — except the vertical motion now starts with $u=u_y$ instead of $u=0$, and it may be moving upward before it starts to fall.

A velocity vector U is drawn at angle alpha above the horizontal. Dashed lines show its horizontal component U cos alpha and vertical component U sin alpha, forming a right-angled triangle.
Resolving the launch velocity U into a horizontal component Ucos α and a vertical component Usin α.

Worked example 1

Resolving a velocity given as speed and angle
A particle is projected from a point on horizontal ground with speed $65\ \text{m s}^{-1}$, at an angle $\alpha$ above the horizontal, where $\tan\alpha = \dfrac{5}{12}$. Find, in the form $(a\mathbf{i}+b\mathbf{j})\ \text{m s}^{-1}$, the initial velocity of the particle, taking $\mathbf{i}$ and $\mathbf{j}$ as horizontal and vertical unit vectors.
MethodUse the 5–12–13 triangle to find sinα and cosα
1
$\tan\alpha=\dfrac{5}{12}$ corresponds to a right-angled triangle with opposite $5$, adjacent $12$, hypotenuse $\sqrt{5^2+12^2}=13$.
Recognising the 5–12–13 triple avoids finding $\alpha$ itself.
2
So $\sin\alpha=\dfrac{5}{13}$ and $\cos\alpha=\dfrac{12}{13}$.
Read these straight off the triangle.
3
$u_x = U\cos\alpha = 65\times\dfrac{12}{13} = 60\ \text{m s}^{-1}$.
Horizontal component uses cosine, adjacent to $\alpha$.
4
$u_y = U\sin\alpha = 65\times\dfrac{5}{13} = 25\ \text{m s}^{-1}$.
Vertical component uses sine, opposite $\alpha$.
5
The initial velocity is $(60\mathbf{i}+25\mathbf{j})\ \text{m s}^{-1}$.
Write the two components as a single vector.
▪ solved

Working backwards: from a vector to speed and angle

You'll also meet the reverse problem — given the velocity as a vector, find the launch speed and angle. Since $u_x=U\cos\alpha$ and $u_y=U\sin\alpha$, it follows that $U=\sqrt{u_x^2+u_y^2}$ (Pythagoras) and $\tan\alpha=\dfrac{u_y}{u_x}$ (dividing one equation by the other).

Worked example 2

Finding speed and angle from a velocity vector
A particle is projected with initial velocity $(9\mathbf{i}+40\mathbf{j})\ \text{m s}^{-1}$, where $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertical unit vectors. Find the initial speed of the particle, and the angle of projection above the horizontal, to the nearest degree.
MethodCombine the components with Pythagoras and the tangent ratio
1
Here $u_x=9$ and $u_y=40$.
Read the components directly from the $\mathbf{i}$ and $\mathbf{j}$ coefficients.
2
$U = \sqrt{u_x^2+u_y^2} = \sqrt{81+1600} = \sqrt{1681} = 41\ \text{m s}^{-1}$.
The speed is the magnitude of the velocity vector.
3
$\tan\alpha = \dfrac{u_y}{u_x} = \dfrac{40}{9} \Rightarrow \alpha = 77^\circ$ (nearest degree).
The angle above the horizontal comes from the ratio of vertical to horizontal component.
▪ solved
Recognising a Pythagorean triple

$9,40,41$ is another Pythagorean triple ($9^2+40^2=81+1600=1681=41^2$), which is why $U$ came out as a whole number. Exam questions are often built around these deliberately — it's worth having 3–4–5, 5–12–13 and 9–40–41 in mind.

Common mistake

Mixing up which component uses sine and which uses cosine. If $\alpha$ is measured from the horizontal, the component along the horizontal is always $U\cos\alpha$ and the component perpendicular to it (vertical) is always $U\sin\alpha$ — regardless of how the triangle happens to be drawn.

Check yourself

A particle is projected from a point on horizontal ground with speed $26\ \text{m s}^{-1}$, at an angle $\alpha$ above the horizontal, where $\tan\alpha=\dfrac{5}{12}$. Express the initial velocity in the form $(a\mathbf{i}+b\mathbf{j})\ \text{m s}^{-1}$.

From $\tan\alpha=\dfrac{5}{12}$: $\sin\alpha=\dfrac{5}{13}$, $\cos\alpha=\dfrac{12}{13}$.

$u_x = 26\times\dfrac{12}{13}=24$, $\;u_y=26\times\dfrac{5}{13}=10$, so the velocity is $(24\mathbf{i}+10\mathbf{j})\ \text{m s}^{-1}$.

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