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## Class 11 Physics (India)

### Course: Class 11 Physics (India)>Unit 8

Lesson 5: Horizontally launched projectiles

# What are velocity components?

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.
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, $\theta$, as seen below.
$\mathrm{sin}\phantom{\rule{0.167em}{0ex}}\theta =\frac{\text{opposite}}{\text{hypotenuse}}$
$\mathrm{cos}\phantom{\rule{0.167em}{0ex}}\theta =\frac{\text{adjacent}}{\text{hypotenuse}}$
$\mathrm{tan}\phantom{\rule{0.167em}{0ex}}\theta =\frac{\text{opposite}}{\text{adjacent}}$

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.
$\mathrm{sin}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{y}}{v}$
$\mathrm{cos}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{x}}{v}$
$\mathrm{tan}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{y}}{{v}_{x}}$

Note that the $v$s in these formulas refer to the magnitudes of the total velocity vector, the total speed, and can therefore never be negative. The individual components ${v}_{x}$ and ${v}_{y}$ can be negative if they point in a negative direction. The convention is that left is negative for the horizontal direction, $x$, and down is negative for the vertical direction, $y$.

## 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 $\theta$ 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 $\text{tan}\theta$.
$\mathrm{tan}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{y}}{{v}_{x}}$
By taking inverse tangent, we get the angle of the total velocity vector in terms of the components.
$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)$

## What's confusing about vector components?

When using $\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)$, 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.
So, for example, if the components of a vector are and , the vector must point leftward—because ${v}_{x}$ is negative—and up—because ${v}_{y}$ is positive.
Concept check: If a paper airplane has the velocity components and , which direction is the paper airplane moving—assuming we choose right and up as positive directions?

## 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 $sin\theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{{v}_{y}}{v}$. The hypotenuse is the magnitude of the velocity 24.3 m/s, $v$, and the opposite side to the angle 30${}^{\circ }$ is ${v}_{y}$.
$\mathrm{sin}\theta =\frac{{v}_{y}}{v}\phantom{\rule{2em}{0ex}}\text{(Use the definition of sine.)}$
${v}_{y}=v\mathrm{sin}\theta \phantom{\rule{2em}{0ex}}\text{(Solve for vertical component.)}$
To find the horizontal component, we'll use $\mathrm{cos}\theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{v}_{x}}{v}$.
$\mathrm{cos}\theta =\frac{{v}_{x}}{v}\phantom{\rule{2em}{0ex}}\text{(Use the definition of cosine.)}$
${v}_{x}=v\mathrm{cos}\theta \phantom{\rule{2em}{0ex}}\text{(Solve for horizontal component.)}$

### Example 2: Angry seagull

An angry seagull is flying over Seattle with a horizontal component of velocity and a vertical component of velocity .
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.
${v}^{2}={v}_{x}^{2}+{v}_{y}^{2}\phantom{\rule{2em}{0ex}}\text{(The Pythagorean theorem.)}$
$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\phantom{\rule{2em}{0ex}}\text{(Take square root of both sides.)}$
To find the angle, we'll use the definition of $\text{tangent}$, but since we now know $v$, we could have used $\text{sine}$ or $\text{cosine}$.
$\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}\phantom{\rule{2em}{0ex}}\text{(Use the definition of tangent.)}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)\phantom{\rule{2em}{0ex}}\text{(Inverse tangent of both sides.)}$
$\theta ={30.6}^{\circ }\phantom{\rule{2em}{0ex}}\text{(Calculate and celebrate!)}$
Since the vertical component is , we know the vector is directed down, and since , we know the vector is directed right. So, we will draw the vector in the fourth quadrant.
The seagull is moving at an angle of ${30.6}^{\circ }$ below the horizontal.