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

Lesson 7: Projectiles launched from/to a height

# What is 2D projectile motion?

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.
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.

### 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,
$\mathrm{\Delta }x={v}_{x}t\phantom{\rule{1em}{0ex}}$
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 ${a}_{y}=-9.8\frac{\text{m}}{{\text{s}}^{2}}$. Since the vertical acceleration is constant, we can solve for a vertical variable with one of the four kinematic formulas which are shown below.
$1.\phantom{\rule{1em}{0ex}}{v}_{y}={v}_{0y}+{a}_{y}t$
$2.\phantom{\rule{1em}{0ex}}\mathrm{\Delta }y=\left(\frac{{v}_{y}+{v}_{0y}}{2}\right)t$
$3.\phantom{\rule{1em}{0ex}}\mathrm{\Delta }y={v}_{0y}t+\frac{1}{2}{a}_{y}{t}^{2}$
$4.\phantom{\rule{1em}{0ex}}{v}_{y}^{2}={v}_{0y}^{2}+2{a}_{y}\mathrm{\Delta }y$
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 ($x$ or $y$) 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 ${v}_{0y}=0$ (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 ${v}_{0}=8.31\frac{\text{m}}{\text{s}}$ from the roof of a building of height .
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 $\mathrm{\Delta }x={v}_{x}t$. To solve for time, consider the fact that we know three variables in the vertical direction (, ${v}_{0y}=0$, $a=-9.8\frac{\text{m}}{{\text{s}}^{2}}$).
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.
$\mathrm{\Delta }y={v}_{0y}t+\frac{1}{2}{a}_{y}{t}^{2}\phantom{\rule{1em}{0ex}}\text{(use the vertical kinematic formula that doesn’t include final velocity)}$
$-H=\left(0\right)t+\frac{1}{2}\left(-g\right){t}^{2}\phantom{\rule{1em}{0ex}}\text{(plug in known vertical values}\right)$
$t=\sqrt{\frac{2H}{g}}\phantom{\rule{1em}{0ex}}\text{(solve symbolically for time t}\right)$
Now we need to plug this time $t$ into the equation for the horizontal direction to find horizontal displacement $\mathrm{\Delta }x$.
$\mathrm{\Delta }x={v}_{x}t\phantom{\rule{1em}{0ex}}\text{(use the equation for the horizontal displacement)}$
So the water balloon struck the ground horizontally from the edge of the building.

### Example 2: Pumpkin launched at an angle

An air cannon is used to launch a pumpkin off a cliff of height with an initial speed ${v}_{0}=11.4\frac{\text{m}}{\text{s}}$ at an angle of $\theta ={52.1}^{\circ }$ 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 ${v}_{y}$).
Before we can do that, we'll have to find the components of the initial velocity (${v}_{0x}$ and ${v}_{0y}$) using the definitions of sine and cosine.
$\text{cos}\theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{v}_{0x}}{{v}_{0}}\phantom{\rule{1em}{0ex}}\text{(use the definition of cosine)}$
${v}_{0x}=\left(11.4\frac{\text{m}}{\text{s}}\right)\text{cos}\left({52.1}^{\circ }\right)\phantom{\rule{1em}{0ex}}\text{(plug in numerical values)}$
(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 ${v}_{0x}=7.00\frac{\text{m}}{\text{s}}$ will also be the horizontal component of the final velocity ${v}_{x}=7.00\frac{\text{m}}{\text{s}}$ 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.
$\text{sin}\theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{{v}_{0y}}{{v}_{0}}\phantom{\rule{1em}{0ex}}\text{(use the definition of sine)}$
${v}_{0y}=\left(11.4\frac{\text{m}}{\text{s}}\right)\text{sin}\left({52.1}^{\circ }\right)\phantom{\rule{1em}{0ex}}\text{(plug in numerical values)}$
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 $t$.
${v}_{y}^{2}={v}_{0y}^{2}+2{a}_{y}\mathrm{\Delta }y\phantom{\rule{1em}{0ex}}\text{(use the kinematic formula that doesn’t include time)}$
${v}_{y}^{2}=434\frac{{\text{m}}^{2}}{{\text{s}}^{2}}\phantom{\rule{1em}{0ex}}\text{(calculate)}$
${v}_{y}=±\sqrt{434\frac{{\text{m}}^{2}}{{\text{s}}^{2}}}=±20.8\frac{\text{m}}{\text{s}}\phantom{\rule{1em}{0ex}}\text{(take a square root)}$
${v}_{y}=-20.8\frac{\text{m}}{\text{s}}\phantom{\rule{1em}{0ex}}\text{(choose the negative root since the pumpkin will be heading downwards)}$
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).
${v}^{2}={v}_{x}^{2}+{v}_{y}^{2}\phantom{\rule{1em}{0ex}}\text{(use the Pythagorean theorem)}$
${v}^{2}=\left(7.00\frac{\text{m}}{\text{s}}{\right)}^{2}+\left(-20.8\frac{\text{m}}{\text{s}}{\right)}^{2}\phantom{\rule{1em}{0ex}}\text{(plug in horizontal and vertical components of the final velocity)}$
${v}^{2}=482\frac{{\text{m}}^{2}}{{\text{s}}^{2}}\phantom{\rule{1em}{0ex}}\text{(calculate)}$
$v=21.9\frac{\text{m}}{\text{s}}\phantom{\rule{1em}{0ex}}\text{(take a square root)}$
This speed of $v=21.9\frac{\text{m}}{\text{s}}$ 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 $\varphi$ of the final velocity using the definition of tangent.
$\text{tan}\varphi =\frac{\text{opposite}}{\text{adjacent}}=\frac{{v}_{y}}{{v}_{x}}$
$\text{tan}\varphi =\frac{20.8\frac{\text{m}}{\text{s}}}{7.00\frac{\text{m}}{\text{s}}}$
Now taking inverse tangent of both sides we get,
$ta{n}^{-1}\left(\text{tan}\varphi \right)=ta{n}^{-1}\left(\frac{20.8\frac{\text{m}}{\text{s}}}{7.00\frac{\text{m}}{\text{s}}}\right)$
The left hand side just becomes $\varphi$, and we can find the value of the right hand side by plugging into a calculator to get,
$\varphi ={71.4}^{\circ }$

## Want to join the conversation?

• 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.
• In example 2, why did we use tan^-1 instead of sin^-1 or cos^-1?
• 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.
• 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.
• 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?