Can all of these answers be abstracted by doing all the algebraic manipulations beforehand and just plugging in the variables into one gigantic formula?

Yeah, and it's actually a great way to gain insight into the nature of the thing. When you solve a thing numerically, you just get some number (or a vector, etc.) at the end (and maybe some units). But when you solve the thing symbolically, you can interpret the equation, see clearly what's proportional to what, any algebraic symmetry (functional symmetry, being able to swap variables, so on), you can see patterns or that some other quantity might be hidden in the thing. For example, Maybe the expression for the area of a circle shows up somewhere in the final expression, which can suggest a different derivation or interpretation. There are other benefits, too. In general, whenever you can – that is, whenever it's not prohibitively difficult – you should try to solve the thing symbolically to gain the greatest insight. Also, once you have a general expression for a thing, you've essentially solved that _class_ of problem. Imagine if you had to rederive the Pythagorean theorem every time you wanted to use it instead of just being able to plug the numbers into the formula.

For two-dimensional projectile motion, is there ever an instance where the initial horizontal velocity is not constant?

Not if it's a projectile. But if it's some other type of object, like something with an engine, then sure. We don't call airplanes projectiles, though, right?

I tried to do the pumpkin launch problem on my own, using the tactic that Sal used in the last video where he used time to construct the final velocity, but I couldn't get the answer to match up. I realize it takes much more time to do it that way, but is it even possible?

Yes, it's possible. I did the same thing and then used the initial vertical velocity and time to solve for the final vertical velocity. My time calculation was 3.04 seconds. So 9+3.04(-9.81) = -20.8, the same final vertical velocity they came up with.

"Besides, when the pumpkin hits the ground the acceleration will no longer be ay = –9.8 m/s2 since the collision will cause a huge unknown acceleration. " What exactly is this huge acceleration?

It is when the ground stops the movement of the pumpkin.

Hi everyone! Just wondering if anyone knows a good place to learn about air resistance? I know it said here it's to do with calculus and differential equations, both of which I am still a beginner in, but is there anyway to learn it a bit more simply to get a better understanding of it?

The only thing you need to know about air resistance is that ideally, it is directly proportional to the velocity of the object experiencing the resistance. Therefore f-sub-a = kv where k is a constant. In your differential equations, remember that velocity is the first time derivative of displacement. Note: I'm using the word 'ideally' a bit loosely here.

Why is the acceleration of projectiles always -g and not g? Why is g negative for projectiles?

I agree with Andrew M's answer, but it also depends on what you make in your equation and drawing (if you do a drawing) positive and negative. If you want down and to the right positive then anything going in those directions are positive and opposite negative, if you make left and up positive then right and down negative. You do whatever you want with equations as long as you know how to do it. To answer you question mainly its because projectiles fall so they go down and that's down like acceleration due to gravity....negative and negative make positive. The projectile might even be negative when it is first launched until it reaches max height.

in the second question, if it was projected upwards with some angle should'nt the vertical displacement be more than 18 m?

Yes. If the projectile is launched at an angle, there's a non-zero vertical component of initial velocity, as compared to the zero vertical initial velocity the object would experience if launched at a 0 degree angle. This non-zero initial velocity would cause an increase in the max vertical displacement, because it takes some time for the object to start traveling downwards, and in that time, it's going up.

Main content

What is 2D projectile motion?

Google Classroom

Learn about how things fly through the air.

What's a 2D projectile?

In a fructose induced rage, you decide to throw a lime at an angle through the air. It takes a path through space as shown by the curved, dashed line in the diagram below. The lime in this case is considered to be a two-dimensional projectile since it's flying both vertically and horizontally through the air, and it's only under the influence of gravity.

Yes. All objects moving through the air experience some amount of air resistance. However, dealing with air resistance can be tricky, so we'll typically assume that the force of air resistance is negligible (i.e., small enough to be ignored). This is not always a good approximation, but it works really well for objects whose weights are large compared to the force of air resistance.

Keep in mind that the force of air resistance on an object gets larger and larger as the speed of the object increases. So we'll typically consider situations where the speed of an object is small enough that air resistance is still negligible.

Since the gravitational force pulls downward, gravity will only affect the vertical component of the velocity

v_{y}

of the lime. The horizontal component of the velocity

v_{x}

will remain unaffected and stay constant as the lime moves along its path.

Try sliding the dot in the diagram below to see that the vertical velocity

v_{y}

changes, but the horizontal velocity

v_{x}

remains constant.

Concept check: At the maximum height of the lime's trajectory, what is the value of the vertical component of velocity?

How do we handle 2D projectile motion mathematically?

One of the easiest ways to deal with 2D projectile motion is to just analyze the motion in each direction separately. In other words, we will use one set of equations to describe the horizontal motion of the lime, and another set of equations to describe the vertical motion of the lime. This turns a single difficult 2D problem into two simpler 1D problems. We're able to do this since the change in the vertical velocity of the lime does not affect the horizontal velocity of the lime. Similarly, throwing the lime with a large horizontal velocity does not affect the vertical acceleration of the lime. In other words, if you fire a bullet horizontally and drop a bullet at the same time, they will hit the ground at the same time.

Yes. This idea is admittedly very counterintuitive to many people, but if air resistance is negligible, and the bullet is fired exactly horizontally, then the shot bullet will hit the ground at the same time as the dropped bullet.

The reason is that the horizontal velocity does not affect the vertical acceleration of the bullet, so both bullets change their vertical velocity by the same amount in the same amount of time (i.e., the fact that one bullet has horizontal velocity does not change the rate at which it moves downward).

Horizontal direction:

There's no acceleration in the horizontal direction since gravity does not pull projectiles sideways, only downward. Air resistance would cause a horizontal acceleration, slowing the horizontal motion, but since we're going to only consider cases where air resistance is negligible we can assume that the horizontal velocity is constant for a projectile.

So for the horizontal direction we can use the following equation,

Δ x = v_{x} t

Whenever the velocity is constant in a certain direction we can use

Δ x = v t

to find how far the object moves in that direction.

Another way to think about it is that constant horizontal velocity means

a_{x} = 0

. So if we plug

a_{x} = 0

into the kinematic formula

Δ x = v_{0 x} t + \frac{1}{2} a_{x} t^{2}

we simply get

Δ x = v_{0 x} t

. And we don't really need to call it the initial horizontal velocity since the horizontal velocity is constant. Replacing

v_{0 x}

with just simply

v_{x}

we again just get

Δ x = v_{x} t

Note: Be sure to only plug horizontal variables into this horizontal equation. If we know two of the variables in this equation we can solve for the remaining unknown variable.

Vertical direction:

Two-dimensional projectiles experience a constant downward acceleration due to gravity

a_{y} = - 9.8 \frac{m}{s^{2}}

. Since the vertical acceleration is constant, we can solve for a vertical variable with one of the four kinematic formulas which are shown below.

1. v_{y} = v_{0 y} + a_{y} t

2. Δ y = (\frac{v_{y} + v_{0 y}}{2}) t

3. Δ y = v_{0 y} t + \frac{1}{2} a_{y} t^{2}

4. v_{y}^{2} = v_{0 y}^{2} + 2 a_{y} Δ y

Be sure to only plug vertical variables into these vertical equations. If we know three of the variables in these equations we can solve for any of the remaining unknown variables.

Note: For a given process, the time interval

t

has the same value for the vertical and horizontal equations. This means that if we ever solve for the time

t

, we can plug that time

t

into the equations for either the vertical or horizontal directions. This strategy is used in many problems. Often, one has to solve for time

t

using the vertical equations, then plug that time into the horizontal equation (or vice versa).

What's confusing about 2D projectile motion?

Many times, people try to substitute vertical components into a horizontal equation, or vice versa. Analyzing each direction (horizontal and vertical) of a projectile independently only works if you keep the different directions (

x

y

) in their own separate equations.

Initial velocities that are directed diagonally will have to be broken into vertical and horizontal components. People sometimes have a hard time breaking a velocity vector into vertical and horizontal components. See this article for help with the trigonometry you use to break vectors into components.

When a projectile is shot horizontally, the initial vertical velocity is zero

v_{0 y} = 0

(see example 1 below). Many learners have a hard time understanding that an object can start with a horizontal component of velocity, yet have zero vertical component of velocity.

What do solved examples involving 2D projectile motion look like?

Example 1: Horizontally launched water balloon

A water balloon is thrown horizontally with a speed of

v_{0} = 8.31 \frac{m}{s}

from the roof of a building of height

H = 23.0 m

How far does the balloon travel horizontally before striking the ground?

We can start by drawing a diagram that includes the given variables.

Once we find the time of flight

t

, we'll be able to solve for the horizontal displacement using

Δ x = v_{x} t

. To solve for time, consider the fact that we know three variables in the vertical direction (

Δ y = - 23.0 m

v_{0 y} = 0

a = - 9.8 \frac{m}{s^{2}}

The vertical component of the initial velocity is zero (

v_{0 y} = 0

) since the balloon was thrown horizontally to start. In order to have a vertical component of initial velocity the balloon would have to be thrown diagonally at an angle upward or downward to start.

The vertical displacement is

- 23 m

since the balloon fell downward through the entire height of the building.

The acceleration in the vertical direction for a projectile is always the acceleration due to gravity

a_{y} = - g = - 9.8 \frac{m}{s^{2}}

So we'll use a kinematic formula in the vertical direction to solve for time

t

. We don't know the final velocity

v_{y}

, and we aren't asked for the final velocity

v_{y}

so we'll use the vertical kinematic formula that doesn't include final velocity.

Δ y = v_{0 y} t + \frac{1}{2} a_{y} t^{2} (use the vertical kinematic formula that doesn’t include final velocity)

- H = (0) t + \frac{1}{2} (- g) t^{2} (plug in known vertical values)

t = \sqrt{\frac{2 H}{g}} (solve symbolically for time t)

t = \sqrt{\frac{2 (- 23.0 m)}{- 9.8 \frac{m}{s^{2}}}} = 2.17 s (plug in numerical values and find the time of flight)

Now we need to plug this time

t

into the equation for the horizontal direction to find horizontal displacement

Δ x

Δ x = v_{x} t (use the equation for the horizontal displacement)

Δ x = (8.31 \frac{m}{s}) (2.17 s) (plug in the time of flight and v_{x})

We found the time of flight

t = 2.17 s

in the calculation above using the vertical equations.

The magnitude of the horizontal component of the initial velocity is equal to the entire initial speed

8.31 \frac{m}{s}

since the water balloon was thrown horizontally to start. If the water balloon were thrown at an angle upward or downward we'd have to use trigonometry to find the horizontal component of the initial velocity.

Δ x = 18.0 m (calculate and celebrate)

So the water balloon struck the ground

18.0 m

horizontally from the edge of the building.

Yes, and solving problems symbolically is awesome because by doing so you essentially solve every possible problem of that type.

For instance, if we plug the symbolic expression for the time

t = \sqrt{\frac{2 H}{g}}

and initial velocity

v_{0}

into the formula for the horizontal displacement

Δ x = v_{x} t

, we'll get the formula below,

Δ x = v_{0} \sqrt{\frac{2 H}{g}}

This formula will work for any projectile thrown horizontally at any velocity

v_{0}

from the edge of a building of any height

H

. We just need to plug in the height and initial velocity for our particular problem and this formula will give us the answer for how far the projectile travels horizontally before striking the ground.

Example 2: Pumpkin launched at an angle

An air cannon is used to launch a pumpkin off a cliff of height

H = 18.0 m

with an initial speed

v_{0} = 11.4 \frac{m}{s}

at an angle of

θ = {52.1}^{\circ}

as seen in the diagram below.

What is the speed of the pumpkin right before it hits the ground?

We'll be able to determine the final speed of the pumpkin if we can determine the components of the final velocity (

v_{x}

and

v_{y}

We're looking for the speed of the pumpkin right before it hits the ground, so we will not assume the final velocity is zero.

Besides, when the pumpkin hits the ground the acceleration will no longer be

a_{y} = - 9.8 \frac{m}{s^{2}}

since the collision will cause a huge unknown acceleration. The kinematic formulas are only true for time intervals where the acceleration is constant. This means we usually don't include the time when the projectile hits the ground, since we typically don't have information about the acceleration during that time.

Before we can do that, we'll have to find the components of the initial velocity (

v_{0 x}

and

v_{0 y}

) using the definitions of sine and cosine.

cos θ = \frac{adjacent}{hypotenuse} = \frac{v_{0 x}}{v_{0}} (use the definition of cosine)

v_{0 x} = v_{0} cos θ (solve for v_{0 x})

v_{0 x} = (11.4 \frac{m}{s}) cos ({52.1}^{\circ}) (plug in numerical values)

v_{0 x} = 7.00 \frac{m}{s} (calculate to find v_{0 x})

(Note: If that seemed like indecipherable mathematical witchcraft, check out this article for help with breaking vectors into components.)

This value we found for the horizontal component of the initial velocity

v_{0 x} = 7.00 \frac{m}{s}

will also be the horizontal component of the final velocity

v_{x} = 7.00 \frac{m}{s}

since the horizontal component of the velocity remains constant for the entire flight (assuming there is no air resistance).

To find the vertical component of the initial velocity we'll use the same procedure as above but with sine instead of cosine.

sin θ = \frac{opposite}{hypotenuse} = \frac{v_{0 y}}{v_{0}} (use the definition of sine)

v_{0 y} = v_{0} sin θ (solve for v_{0 y})

v_{0 y} = (11.4 \frac{m}{s}) sin ({52.1}^{\circ}) (plug in numerical values)

v_{0 y} = 9.00 \frac{m}{s} (calculate to find v_{0 y})

Since the vertical component of velocity

v_{y}

changes for a projectile as it moves through the air we'll have to solve for the vertical component of the final velocity

v_{y}

using a kinematic formula for the vertical direction. Since we don't know the time of flight

t

, and we weren't asked to find the time

t

, we'll use the vertical kinematic formula that doesn't include time

t

v_{y}^{2} = v_{0 y}^{2} + 2 a_{y} Δ y (use the kinematic formula that doesn’t include time)

v_{y}^{2} = (9.00 \frac{m}{s})^{2} + 2 (- 9.8 \frac{m}{s^{2}}) (- 18 m) (plug in known values)

v_{y}^{2} = 434 \frac{m^{2}}{s^{2}} (calculate)

v_{y} = \pm \sqrt{434 \frac{m^{2}}{s^{2}}} = \pm 20.8 \frac{m}{s} (take a square root)

v_{y} = - 20.8 \frac{m}{s} (choose the negative root since the pumpkin will be heading downwards)

Now that we know the horizontal and vertical components of the final velocity, we can use the Pythagorean theorem to solve for the final speed (i.e. magnitude of the final velocity).

v^{2} = v_{x}^{2} + v_{y}^{2} (use the Pythagorean theorem)

v^{2} = (7.00 \frac{m}{s})^{2} + (- 20.8 \frac{m}{s})^{2} (plug in horizontal and vertical components of the final velocity)

v^{2} = 482 \frac{m^{2}}{s^{2}} (calculate)

v = 21.9 \frac{m}{s} (take a square root)

This speed of

v = 21.9 \frac{m}{s}

is the magnitude of the final velocity of the pumpkin right before it hits the ground. The relationship between the final velocity and its components is shown in the diagram below.

We could also solve for the angle

ϕ

of the final velocity using the definition of tangent.

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

tan ϕ = \frac{20.8 \frac{m}{s}}{7.00 \frac{m}{s}}

We're just going to solve for the magnitude of the angle between the horizontal and the final velocity so we're just going to plug in the magnitudes of the horizontal and vertical velocities.

The reason we do this is that the calculator often can't tell whether the negative came from the numerator or the denominator of the tangent, which makes it a little unreliable in letting us know which quadrant the angle lies. To avoid this negative sign issue we can just draw which way the final velocity points, treat the sides of the triangle/components of the velocity as positive, and then find the magnitude of the desired angle.

Now taking inverse tangent of both sides we get,

t a n^{- 1} (tan ϕ) = t a n^{- 1} (\frac{20.8 \frac{m}{s}}{7.00 \frac{m}{s}})

The left hand side just becomes

ϕ

, and we can find the value of the right hand side by plugging into a calculator to get,

ϕ = {71.4}^{\circ}

Want to join the conversation?

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Adam
Posted 9 years ago. Direct link to Adam's post “Can all of these answers ...”
Can all of these answers be abstracted by doing all the algebraic manipulations beforehand and just plugging in the variables into one gigantic formula?
Button navigates to signup pageComment on Adam's post “Can all of these answers ...”
(52 votes)
Answer
- Alia
  Posted a year ago. Direct link to Alia's post “Yeah, and it's actually a...”
  Yeah, and it's actually a great way to gain insight into the nature of the thing. When you solve a thing numerically, you just get some number (or a vector, etc.) at the end (and maybe some units). But when you solve the thing symbolically, you can interpret the equation, see clearly what's proportional to what, any algebraic symmetry (functional symmetry, being able to swap variables, so on), you can see patterns or that some other quantity might be hidden in the thing. For example, Maybe the expression for the area of a circle shows up somewhere in the final expression, which can suggest a different derivation or interpretation.
  
  There are other benefits, too. In general, whenever you can – that is, whenever it's not prohibitively difficult – you should try to solve the thing symbolically to gain the greatest insight. Also, once you have a general expression for a thing, you've essentially solved that class of problem. Imagine if you had to rederive the Pythagorean theorem every time you wanted to use it instead of just being able to plug the numbers into the formula.
  Button navigates to signup page
  (5 votes)
shreynathan
Posted 7 years ago. Direct link to shreynathan's post “For two-dimensional proje...”
For two-dimensional projectile motion, is there ever an instance where the initial horizontal velocity is not constant?
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(7 votes)
Answer
- Andrew M
  Posted 7 years ago. Direct link to Andrew M's post “Not if it's a projectile....”
  Not if it's a projectile. But if it's some other type of object, like something with an engine, then sure. We don't call airplanes projectiles, though, right?
  Comment on Andrew M's post “Not if it's a projectile....”
  (8 votes)
Kevin Niemeyer
Posted 8 years ago. Direct link to Kevin Niemeyer's post “I tried to do the pumpkin...”
I tried to do the pumpkin launch problem on my own, using the tactic that Sal used in the last video where he used time to construct the final velocity, but I couldn't get the answer to match up. I realize it takes much more time to do it that way, but is it even possible?
Button navigates to signup pageComment on Kevin Niemeyer's post “I tried to do the pumpkin...”
(4 votes)
Answer
- Zak.Witter
  Posted 8 years ago. Direct link to Zak.Witter's post “Yes, it's possible. I did...”
  Yes, it's possible. I did the same thing and then used the initial vertical velocity and time to solve for the final vertical velocity. My time calculation was 3.04 seconds. So 9+3.04(-9.81) = -20.8, the same final vertical velocity they came up with.
  Button navigates to signup page
  (7 votes)
David
Posted 7 years ago. Direct link to David's post “"Besides, when the pumpki...”
"Besides, when the pumpkin hits the ground the acceleration will no longer be ay = –9.8 m/s2 since the collision will cause a huge unknown acceleration. "

What exactly is this huge acceleration?
Button navigates to signup pageComment on David's post “"Besides, when the pumpki...”
(2 votes)
Answer
- Charles LaCour
  Posted 7 years ago. Direct link to Charles LaCour's post “It is when the ground sto...”
  It is when the ground stops the movement of the pumpkin.
  Comment on Charles LaCour's post “It is when the ground sto...”
  (8 votes)
Danilo Devedžić
Posted 7 years ago. Direct link to Danilo Devedžić's post “In example 2, why did we ...”
In example 2, why did we use tan^-1 instead of sin^-1 or cos^-1?
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(3 votes)
Answer
- Vuk Perović
  Posted 4 years ago. Direct link to Vuk Perović's post “We could've used those to...”
  We could've used those too, you're right. They might have wanted to use a trigonometric function which they hadn't used before in that example.
  Button navigates to signup page
  (0 votes)
Toska Pi
Posted 5 years ago. Direct link to Toska Pi's post “Hi everyone! Just wonderi...”
Hi everyone! Just wondering if anyone knows a good place to learn about air resistance? I know it said here it's to do with calculus and differential equations, both of which I am still a beginner in, but is there anyway to learn it a bit more simply to get a better understanding of it?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- The #1 Pokemon Proponent
  Posted 5 years ago. Direct link to The #1 Pokemon Proponent's post “The only thing you need t...”
  The only thing you need to know about air resistance is that ideally, it is directly proportional to the velocity of the object experiencing the resistance.
  Therefore f-sub-a = kv where k is a constant. In your differential equations, remember that velocity is the first time derivative of displacement.
  Note: I'm using the word 'ideally' a bit loosely here.
  Comment on The #1 Pokemon Proponent's post “The only thing you need t...”
  (4 votes)
Sarah
Posted 4 years ago. Direct link to Sarah's post “For example 1, isn't the ...”
For example 1, isn't the equation for horizontal displacement s=((v+u)t)/2? Which would give you a displacement of 9 meters, instead of 18.0m?
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(3 votes)
Answer
veronicasih18
Posted 7 years ago. Direct link to veronicasih18's post “Why is the acceleration o...”
Why is the acceleration of projectiles always -g and not g? Why is g negative for projectiles?
Button navigates to signup pageComment on veronicasih18's post “Why is the acceleration o...”
(2 votes)
Answer
- Carl Daoud
  Posted 7 years ago. Direct link to Carl Daoud's post “I agree with Andrew M's a...”
  I agree with Andrew M's answer, but it also depends on what you make in your equation and drawing (if you do a drawing) positive and negative. If you want down and to the right positive then anything going in those directions are positive and opposite negative, if you make left and up positive then right and down negative. You do whatever you want with equations as long as you know how to do it. To answer you question mainly its because projectiles fall so they go down and that's down like acceleration due to gravity....negative and negative make positive. The projectile might even be negative when it is first launched until it reaches max height.
  Button navigates to signup page
  (2 votes)
Cindy
Posted 8 years ago. Direct link to Cindy's post “For the first example wit...”
For the first example with the water balloon, they found the flight time using the vertical kinematic formulas, but then used the vertical flight time and plugged it into the horizontal formula. I thought you weren't supposed to put vertical values into horizontal formulas and vice versa. Can someone please explain this to me?
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(3 votes)
Answer
- The #1 Pokemon Proponent
  Posted 5 years ago. Direct link to The #1 Pokemon Proponent's post “First of all, did you und...”
  First of all, did you understand what that time represents? It's the time the object was in the air. Yes it was moving vertically, but also horizontally and it stops moving horizontally the instant it hits the ground.(do not assume it bounces unless it is mentioned). Therefore the flight time is also the total time the object is moving horizontally. The object has no horizontal acceleration and hence you can just use d=v*t where t is the time of flight, v is the initial horizontal velocity(it should continue for the whole motion) and d is the distance it covers horizontally.
  Button navigates to signup page
  (1 vote)
aayushbirijal1
Posted 4 years ago. Direct link to aayushbirijal1's post “in the second question, i...”
in the second question, if it was projected upwards with some angle should'nt the vertical displacement be more than 18 m?
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(1 vote)
Answer
- Hecretary Bird
  Posted 4 years ago. Direct link to Hecretary Bird's post “Yes. If the projectile is...”
  Yes. If the projectile is launched at an angle, there's a non-zero vertical component of initial velocity, as compared to the zero vertical initial velocity the object would experience if launched at a 0 degree angle. This non-zero initial velocity would cause an increase in the max vertical displacement, because it takes some time for the object to start traveling downwards, and in that time, it's going up.
  Comment on Hecretary Bird's post “Yes. If the projectile is...”
  (2 votes)

Physics library

Course: Physics library > Unit 2