What are velocity components? (article)

Learn how to simplify vectors by breaking them into parts.

Why do we break up vectors into components?

Two-dimensional motion is more complex than one-dimensional motion since the velocities can point in diagonal directions. For example, a baseball could be moving both horizontally and vertically at the same time with a diagonal velocity $v$ ‍. We break up the velocity vector, $v$ ‍, of the baseball into two separate horizontal, $v_{x}$ ‍, and vertical, $v_{y}$ ‍, directions to simplify our calculations.

Trying to tackle both the horizontal and vertical directions of a baseball in one single equation is difficult; it’s better to take a divide-and-conquer approach.

Yes. We can break a vector into any two non colinear directions. For example, the vector $v$ ‍ could be broken up into the two directions shown below. We could even choose two directions that aren't perpendicular, which turns out to be useful in special relativity.

That said, it's almost always simplest to break vectors into perpendicular components. And it's often most convenient to choose horizontal and vertical components since the acceleration due to gravity lies entirely along the vertical direction—in other words, straight down.

Also, many typical velocity questions are explicitly horizontal or vertical in nature and are therefore best solved for using horizontal and vertical axes. For example: "How far horizontally does the projectile make it?" or "How high does the projectile make it?"

Breaking up the diagonal velocity $v$ ‍ into horizontal $v_{x}$ ‍ and vertical $v_{y}$ ‍ components allows us to deal with each direction separately. Essentially, we'll be able to turn one difficult two-dimensional problem into two easier one-dimensional problems. This trick of breaking up vectors into components works even when the vector is something other than velocity, for example, forces, momentum, or electric fields. In fact, you'll use this trick over and over in physics, so it's important to get really good at dealing with vector components as soon as possible.

How do we break a vector into components?

Before we talk about breaking up vectors, we should note that trigonometry already gives us the ability to relate the side lengths of a right triangle—hypotenuse, opposite, adjacent—and one of the angles, $θ$ ‍, as seen below.

$\sin θ = \frac{opposite}{hypotenuse}$ ‍

$\cos θ = \frac{adjacent}{hypotenuse}$ ‍

$\tan θ = \frac{opposite}{adjacent}$ ‍

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When we break any diagonal vector into two perpendicular components, the total vector and its components— $v, v_{y}, v_{x}$ ‍—form a right triangle. Because of this, we can apply the same trigonometric rules to a velocity vector magnitude and its components, as seen below. Notice that $v_{x}$ ‍ is treated as the adjacent side, $v_{y}$ ‍ as the opposite, and $v$ ‍ as the hypotenuse.

$\sin θ = \frac{v_{y}}{v}$ ‍

$\cos θ = \frac{v_{x}}{v}$ ‍

$\tan θ = \frac{v_{y}}{v_{x}}$ ‍

If we change the angle under consideration to the other angle—the angle formed by the vertical axis—the rules of trigonometry stay the same, but what we used to call the opposite side will become the adjacent side and vice versa.

$\sin ϕ = \frac{opposite}{hypotenuse} = \frac{v_{x}}{v}$ ‍

$\cos ϕ = \frac{adjacent}{hypotenuse} = \frac{v_{y}}{v}$ ‍

$\tan ϕ = \frac{opposite}{adjacent} = \frac{v_{x}}{v_{y}}$ ‍

How do you determine the magnitude and angle of the total vector?

We saw in the previous sections how a vector magnitude and angle can be broken up into vertical and horizontal components. But what if you start with some given velocity components: $v_{y}$ ‍ and $v_{x}$ ‍? How could you use the components to find the magnitude $v$ ‍ and angle $θ$ ‍ of the total velocity vector?

Finding the magnitude of the total velocity vector isn't too hard since for any right triangle the side lengths and hypotenuse will be related by the Pythagorean theorem.

$v^{2} = {v_{x}}^{2} + {v_{y}}^{2}$ ‍

By taking a square root, we get the magnitude of the total velocity vector in terms of the components.

$v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}$ ‍

Also, if we know both components of the total vector, we can find the angle of the total vector using $tan θ$ ‍.

$\tan θ = \frac{v_{y}}{v_{x}}$ ‍

By taking inverse tangent, we get the angle of the total velocity vector in terms of the components.

$θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})$ ‍

Since the definition of $\tan$ ‍—opposite over adjacent or, in this case, $v_{y}$ ‍ over $v_{x}$ ‍—only includes the components of the vector, it conveniently allows us to find the angle $θ$ ‍ of the vector without knowing the magnitude of the total vector $v$ ‍.

If, however, we already know the magnitude of the total vector $v$ ‍ and the components $v_{y}$ ‍ and $v_{x}$ ‍, we can use sine, cosine, or tangent to find the angle of the total vector.

What's confusing about vector components?

When using $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})$ ‍, the fact that we put $v_{y}$ ‍ on top as the opposite side and the $v_{x}$ ‍ on the bottom as the adjacent side means that we are measuring the angle from the horizontal axis. It seems like figuring out how to draw the angle could be confusing, but here are two good tips:

Assuming we have selected right/up as the positive directions, if the horizontal component $v_{x}$ ‍ is positive, the vector points rightward. If the horizontal component, $v_{x}$ ‍ is negative, the vector points leftward.

Again, asumming we have selected right/up as the positive directions, if the vertical component $v_{y}$ ‍ is positive, the vector points upward. If the vertical component $v_{y}$ ‍ is negative, the vector points downward.

What do solved examples involving vector components look like?

Example 1: Bend it like Beckham

A soccer ball is kicked up and to the right at an angle of 30 $^{\circ}$ ‍ with a speed of 24.3 m/s as seen below.

What is the vertical component of the velocity at the moment shown?

What is the horizontal component of the velocity at the moment shown?

To find the vertical component of the velocity, we'll use $s i n θ = \frac{opposite}{hypotenuse} = \frac{v_{y}}{v}$ ‍. The hypotenuse is the magnitude of the velocity 24.3 m/s, $v$ ‍, and the opposite side to the angle 30 $^{\circ}$ ‍ is $v_{y}$ ‍.

$\sin θ = \frac{v_{y}}{v} (Use the definition of sine.)$ ‍

$v_{y} = v \sin θ (Solve for vertical component.)$ ‍

$v_{y} = (24.3 m/s) \sin (30^{\circ}) (Plug in values.)$ ‍

$v_{y} = 12.2 m/s (Calculate and celebrate!)$ ‍

To find the horizontal component, we'll use $\cos θ = \frac{adjacent}{hypotenuse} = \frac{v_{x}}{v}$ ‍.

$\cos θ = \frac{v_{x}}{v} (Use the definition of cosine.)$ ‍

$v_{x} = v \cos θ (Solve for horizontal component.)$ ‍

$v_{x} = (24.3 m/s) \cos (30^{\circ}) (Plug in values.)$ ‍

$v_{x} = 21.0 m/s (Calculate and celebrate!)$ ‍

Example 2: Angry seagull

An angry seagull is flying over Seattle with a horizontal component of velocity $v_{x} = 14.6 m/s$ ‍ and a vertical component of velocity $v_{y} = - 8.62 m/s$ ‍.

What is the magnitude of the total velocity of the seagull?

What is the angle of the total velocity?

Assume right/up are positive, and assume all angles will be measured counterclockwise from the positive x axis.

We'll use the Pythagorean theorem to find the magnitude of the total velocity vector.

$v^{2} = v_{x}^{2} + v_{y}^{2} (The Pythagorean theorem.)$ ‍

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} (Take square root of both sides.)$ ‍

$v = \sqrt{(14.6 m/s)^{2} + (- 8.62 m/s)^{2}} (Plug in.)$ ‍

$v = 17.0 m/s (Calculate and celebrate!)$ ‍

To find the angle, we'll use the definition of $tangent$ ‍, but since we now know $v$ ‍, we could have used $sine$ ‍ or $cosine$ ‍.

$\tan θ = \frac{v_{y}}{v_{x}} (Use the definition of tangent.)$ ‍

$θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) (Inverse tangent of both sides.)$ ‍

$θ = \tan^{- 1} (\frac{8.62 m/s}{14.6 m/s}) (Plug in magnitudes.)$ ‍

$θ = {30.6}^{\circ} (Calculate and celebrate!)$ ‍

Since the vertical component is $v_{y} = - 8.62 m/s$ ‍, we know the vector is directed down, and since $v_{x} = 14.6 m/s$ ‍, we know the vector is directed right. So, we will draw the vector in the fourth quadrant.

The seagull is moving $17.0 m/s$ ‍ at an angle of ${30.6}^{\circ}$ ‍ below the horizontal.

What are velocity components? (article) | Khan Academy (2025)