What is the meaning of terminal velocity?

How do we calculate the air resistance and in turn terminal velocity?

Although the same question has been asked, i didn't quite get your conclusion. When finding the angle of the vector, you didn't keep the y-component negative. i did and my angle was negative. Does it matter? My answer's still correct?

In that part of the Angry Seagull question he isn't taking the exact velocities, he is taking the magnitudes of the velocities (just the number, not the sign). Magnitudes are always positive. The angle he gets is positive, because it's the magnitude of that angle. If you were to look at it on a polar graph, you could indeed list that as a negative angle, and it would indicate that the bird is flying right (positive) and down (negative). To sum it all up, if you want to succinctly tell what direction the bird is heading, the negative angle is the correct answer [on a polar graph this would read (17.0 , -30.6 degrees)], but if all you care about is the size of the angle, the answer in the article is correct.

But how would we know if our triangle is right angled? would we have to make a scale diagram first or?? im so confused

Assuming you draw a horizontal and a vertical vector, as they showed, you will always end up with a right triangle. This is because the horizontal and vertical lines are perpendicular to each other, creating a right angle. Of course, you can instead draw the vectors as non-horizontal and vertical lines (as they also showed), but I think it would make the triangle much harder to use. Just remember that as long as your triangle has two lines perpendicular to each other, it is a right triangle with a right angle. Hope this helps! Let me know if you still don't understand =)

so this is kind of like finding the degrees of a point on the unit circle using: sin( ) cos( ) tan( ) csc( ) sec( ) cot( ) arcsin( ) arccos( ) arctan( ) arccsc( ) arcsec( ) arccot( ) Right? Or no?

Sorta? But you will not need to know the list after tangent from what I have read up on the MCAT.

Can I have more than two vectors that affects my velocity?

Yes. There is a 3rd axis of movement known as z used in 3-D diagrams. Movement on this axis implies going in or out of the page. Also you can have an infinite quantity of vectors affecting velocity as long as you account for each individual vector's velocity.

I notice in the presentations, you use double lines on each side of the components. What do the double lines mean?

Hello Bob, When we work with vectors double lines such as this ||X|| or single lines such as |X| are referring to the length of vector X. I hope this is the answer to your question as I did not see this nomenclature in this link. Please leave a comment below. Regards, APD

Can any one give me the all formulas of Projectile motion chapter

http://formulas.tutorvista.com/physics/projectile-motion-formula.html

How do we solve for the components if we are only given the length in which the 2 directions are traveled rather than the velocity?

You need time if you want to solve for velocity components. Maybe you want to solve for displacement components?

How to use inverse trig functions without a calculator?

You should know the trig values for these common angles (0, 30, 45, 60, 90, 120, etc...) which can be expressed with algebraic numbers (ie. roots of integers). If you know the trig values, you also know the inverse trig values as well.

Is arcsin the same as sin^-1 ? And for the rest too?

sin^-1(x) is sometimes used instead of arcsin(x) but if you ask me it is a terrible notation because it can be easily confused with 1/sin(x) which is definitely not arcsin(x). So my suggestion is never use sin^-1(x).

Main content

Course: Class 11 Physics (India) > Unit 8

Lesson 5: Horizontally launched projectiles

What are velocity components?

Google Classroom

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}

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}}

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

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.

So, for example, if the components of a vector are

v_{x} = - 12 m/s

and

v_{y} = 10 m/s

, 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

v_{x} = - 7 m/s

and

v_{y} = - 5 m/s

, which direction is the paper airplane moving—assuming we choose right and up as positive directions?

Not always. The problem is that when we plug negative components into

θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})

, many calculators don't indicate if the negative sign is for the top or bottom component. Also, if we plug in both components as negative, the calculator cancels the negatives and gives an angle as if both components were positive.

This is frustrating, but can be overcome. A useful strategy is to plug in all components into

θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})

as positive numbers. Then, use the rules about component signs to figure out which quadrant to draw the vector in. We’ll use this method in the solved examples section below.

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}

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

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.

Want to join the conversation?

Sort by:

Ama Ojiako
Posted 9 years ago. Direct link to Ama Ojiako's post “Although the same questio...”
Although the same question has been asked, i didn't quite get your conclusion. When finding the angle of the vector, you didn't keep the y-component negative. i did and my angle was negative. Does it matter? My answer's still correct?
Button navigates to signup pageComment on Ama Ojiako's post “Although the same questio...”
(11 votes)
Answer
- r.m.robinson02
  Posted 8 years ago. Direct link to r.m.robinson02's post “In that part of the Angry...”
  In that part of the Angry Seagull question he isn't taking the exact velocities, he is taking the magnitudes of the velocities (just the number, not the sign). Magnitudes are always positive. The angle he gets is positive, because it's the magnitude of that angle. If you were to look at it on a polar graph, you could indeed list that as a negative angle, and it would indicate that the bird is flying right (positive) and down (negative).
  
  To sum it all up, if you want to succinctly tell what direction the bird is heading, the negative angle is the correct answer [on a polar graph this would read (17.0 , -30.6 degrees)], but if all you care about is the size of the angle, the answer in the article is correct.
  Button navigates to signup page
  (26 votes)
Arushi
Posted 8 years ago. Direct link to Arushi's post “What is the meaning of te...”
What is the meaning of terminal velocity?
Button navigates to signup pageComment on Arushi's post “What is the meaning of te...”
(16 votes)
Answer
- Shivansh
  Posted 7 years ago. Direct link to Shivansh's post “How do we calculate the a...”
  How do we calculate the air resistance and in turn terminal velocity?
  Button navigates to signup page
  (3 votes)
carla castaneda
Posted 9 years ago. Direct link to carla castaneda's post “But how would we know if ...”
But how would we know if our triangle is right angled? would we have to make a scale diagram first or?? im so confused
Button navigates to signup pageComment on carla castaneda's post “But how would we know if ...”
(3 votes)
Answer
- Thessalonika
  Posted 8 years ago. Direct link to Thessalonika's post “Assuming you draw a horiz...”
  Assuming you draw a horizontal and a vertical vector, as they showed, you will always end up with a right triangle. This is because the horizontal and vertical lines are perpendicular to each other, creating a right angle. Of course, you can instead draw the vectors as non-horizontal and vertical lines (as they also showed), but I think it would make the triangle much harder to use. Just remember that as long as your triangle has two lines perpendicular to each other, it is a right triangle with a right angle.
  Hope this helps! Let me know if you still don't understand =)
  Comment on Thessalonika's post “Assuming you draw a horiz...”
  (10 votes)
Bob Everton
Posted 7 years ago. Direct link to Bob Everton's post “I notice in the presentat...”
I notice in the presentations, you use double lines on each side of the components. What do the double lines mean?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- APDahlen
  Posted 7 years ago. Direct link to APDahlen's post “Hello Bob, When we work ...”
  Hello Bob,
  
  When we work with vectors double lines such as this ||X|| or single lines such as |X| are referring to the length of vector X.
  
  I hope this is the answer to your question as I did not see this nomenclature in this link. Please leave a comment below.
  
  Regards,
  
  APD
  Button navigates to signup page
  (11 votes)
Michael Headrick
Posted 8 years ago. Direct link to Michael Headrick's post “so this is kind of like f...”
so this is kind of like finding the degrees of a point on the unit circle using:
sin( )
cos( )
tan( )
csc( )
sec( )
cot( )
arcsin( )
arccos( )
arctan( )
arccsc( )
arcsec( )
arccot( )

Right?
Or no?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- rerostch
  Posted 8 years ago. Direct link to rerostch's post “Sorta? But you will not n...”
  Sorta? But you will not need to know the list after tangent from what I have read up on the MCAT.
  Button navigates to signup page
  (6 votes)
A01373541
Posted 9 years ago. Direct link to A01373541's post “Can I have more than two ...”
Can I have more than two vectors that affects my velocity?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Zander Hall
  Posted 9 years ago. Direct link to Zander Hall's post “Yes. There is a 3rd axis ...”
  Yes. There is a 3rd axis of movement known as z used in 3-D diagrams. Movement on this axis implies going in or out of the page. Also you can have an infinite quantity of vectors affecting velocity as long as you account for each individual vector's velocity.
  Button navigates to signup page
  (6 votes)
zbahalkani8
Posted 8 years ago. Direct link to zbahalkani8's post “Can any one give me the a...”
Can any one give me the all formulas of Projectile motion chapter
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Sameer Ahmed
  Posted 8 years ago. Direct link to Sameer Ahmed's post “http://formulas.tutorvist...”
  http://formulas.tutorvista.com/physics/projectile-motion-formula.html
  Button navigates to signup page
  (6 votes)
Victoria
Posted 8 years ago. Direct link to Victoria's post “Is arcsin the same as sin...”
Is arcsin the same as sin^-1 ? And for the rest too?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- Mark Zwald
  Posted 8 years ago. Direct link to Mark Zwald's post “sin^-1(x) is sometimes us...”
  sin^-1(x) is sometimes used instead of arcsin(x) but if you ask me it is a terrible notation because it can be easily confused with 1/sin(x) which is definitely not arcsin(x). So my suggestion is never use sin^-1(x).
  Comment on Mark Zwald's post “sin^-1(x) is sometimes us...”
  (5 votes)
atass.y.417
Posted 8 years ago. Direct link to atass.y.417's post “How do we solve for the c...”
How do we solve for the components if we are only given the length in which the 2 directions are traveled rather than the velocity?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Andrew M
  Posted 8 years ago. Direct link to Andrew M's post “You need time if you want...”
  You need time if you want to solve for velocity components.
  Maybe you want to solve for displacement components?
  Button navigates to signup page
  (3 votes)
Shivam Rawat
Posted 8 years ago. Direct link to Shivam Rawat's post “How to use inverse trig ...”
How to use inverse trig functions without a calculator?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Mark Zwald
  Posted 8 years ago. Direct link to Mark Zwald's post “You should know the trig ...”
  You should know the trig values for these common angles (0, 30, 45, 60, 90, 120, etc...) which can be expressed with algebraic numbers (ie. roots of integers). If you know the trig values, you also know the inverse trig values as well.
  Button navigates to signup page
  (2 votes)