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A scalar magnitude with a direction.

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In order to solve two dimensional problems it is necessary to break all vectors into their x and y components. Different dimensions do not 'talk' to each other. Thus one must use the equations of motion once for the x-direction and once for the y-direction. For example, when working with the x-direction, one only includes the x-component values of the vectors in the calculations. Note that if an object is 'launched horizontally', then the full value is in the x-direction and there is no component in the y-direction.


The first new concept introduced here is that of a vector: a scalar magnitude with a direction. In a sense, we are almost as good at natural vector manipulation as we are at adding numbers. Consider, for instance, throwing a ball to a friend standing some distance away. To perform an accurate throw, one has to figure out both where to throw and how hard. We can represent this concept graphically with an arrow: it has an obvious direction, and its length can represent the distance the ball will travel in a given time. Such a vector (an arrow between the original and final location of an object) is called a displacement:

Vector Components

From the above examples, it should be clear that two vectors add to make another vector. Sometimes, the opposite operation is useful: we often want to represent a vector as the sum of two other vectors. This is called breaking a vector into its components. When vectors point along the same line, they essentially add as scalars. If we break vectors into components along the same lines, we can add them by adding their components. The lines we pick to break our vectors into components along are often called a basis. Any basis will work in the way described above, but we usually break vectors into perpendicular components, since it will frequently allow us to use the Pythagorean theorem in time-saving ways. Specifically, we usually use the \begin{align*} x \end{align*} and \begin{align*} y \end{align*} axes as our basis, and therefore break vectors into what we call their \begin{align*} x \end{align*} and \begin{align*} y \end{align*} components:

A final reason for breaking vectors into perpendicular components is that they are in a sense independent: adding vectors along a component perpendicular to an original component one will never change the original component, just like changing the \begin{align*} y \end{align*}-coordinate of a point can never change its \begin{align*} x \end{align*}-coordinate.

Breaking the Initial Velocity into its Components


Example 1

A tennis ball is launched \begin{align*}32^\circ\end{align*} above the horizontal at a speed of 7.0 m/s. What are the horizontal and vertical velocity components?

Question: \begin{align*}v_x\end{align*} and \begin{align*}v_y = \ ? \ [m/s]\end{align*}

Given\begin{align*}v = 7.0 \ m/s\end{align*}

\begin{align*}{\;} \qquad \quad \theta = 32^\circ\end{align*}

Equations: \begin{align*}v_x = v \cos \theta \qquad v_y = v \sin \theta\end{align*}

Plug n' chug: \begin{align*}v_x = v \cos \theta = (7.0 \ m/s) \cos (32^\circ)=5.9 \ m/s\end{align*}

\begin{align*}{\;}\qquad \qquad \qquad \ v_y = v \sin \theta = (7.0 \ m/s) \sin (32^\circ)= 3.7 \ m/s\end{align*}

The answer is 5.9 m/s, 3.7 m/s.

Interactive Simulation


  1. Find the missing legs or angles of the triangles shown.
  2. Draw in the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}velocity components for each dot along the path of the cannonball. The first one is done for you.

Review (Answers)

1. a. 13m  b. 40.6°  c. vx= 45.03 m/s, vy=26 m/s  d. 56.31°, vx = 16.1 m/s 

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