5. Projectile Motion

Intro to Projectile Motion: Horizontal Launch

1

concept

## Introduction to Projectile Motion

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Hey, guys, for the next couple videos, we're really talking about projectile motion. This is a super important topic in physics. You absolutely will need to know how to solve problems. For now, there's a lot of different variations of problems and project on motion, and it might seem that there's different ways to solve them. So I'm gonna show you in this video. I'm gonna introduce you to the main types of project on motion that you'll see, and I'm gonna show you that to solve them. We're just gonna use equations that we've already seen before, so let's check it out. So project on motion is gonna happen whenever you taken object and you throw it and it moves into dimensions. That's the important part under the influence of Onley gravity. So we've actually seen something similar to this before when we studied one dimensional vertical motion. If you take an object and you throw it straight up, then it comes straight back down on along the same direction that you through it because it's under the influence of only gravity. It just moved purely up and down. What's different about Project on motion is that we're not going to throw something up and down. We're gonna give it some sideways velocity, whether we're throwing it perfectly horizontal, whether we're throwing it downwards or upwards. These air really the main types that you might see, but because we're forgiving, it's, um, sideways velocity. It doesn't just go up and down. Instead, it goes to the side while also falling because, just like in vertical motion, it's still under the influence of only gravity. So because it's moving sideways and up and down, it takes these two dimensional parabolic paths some other examples that you might see or where you throw in something downwards or even upwards. You might see situations where it returns to the same height or might go even to a lower height or even a higher height like this. But this is really it. These are all the different kinds of variations that you might see. All of your problems are gonna fall into one of these categories, so it's common through all of them Is that the acceleration and the only thing that's influencing this object is just gravity that acts vertically downwards. So we have these two dimensional parabolic path. So how do we solve problems or remember whatever. We have physics problems in two dimensions. Whether whether it's motion or vectors, we can always decompose them into one dimensional X and Y parts. It's no different for projectile motion. So we can do here is we can kind of break up this motion this parabola into the X and y axis here. So imagine that we have this object here. Let's call this point a and it moves to the ground, which is Point B. And if we could Onley move along the X axis, then what it would look like is, even though this is taking a parabolic path like this in the X axis, this object just moves from A to B like this. Now it's also moving in the Y axis, so from a it's also falling under the influence of gravity like this. So it's really just doing two of these motions at the same time, and that's why it looks like a parabola. We can actually do this for any other motion. I'm just gonna pick this one over here, so like, for instance, like a and then it goes up to some points. Let's call that be, Then it goes back down again. Let's call that point. C. You can just take this and break it up into its too one dimensional parts in the X and y. So if you Onley could move along the x axis, then in the X axis you just go from a to B to C like this in the y axis. If you'd Onley move along the y axis what this would project? I would look like it would go from a and then it would go up to be And then we go back down to see so would actually look a little bit like this would actually look exactly like this motion over here. So really, it's just these 21 dimensional motions that are happening at the same time. Now, why is this important? Because we take a look here. The Onley acceleration at this object experiences is just gonna be in the Y axis due to gravity. So when it comes to these 21 dimensional motions, projectile motion is really just a combination of horizontal motion where the acceleration in the X axis is equal to zero and vertical motion where the acceleration in the Y axis is equal to negative G. It's negative because we're always gonna be assume that the up into the right is positive and G acts downwards. So it's negative. Now, why is this important? Because when it comes to the equations that we use for project on motion, we're really just gonna be using combinations of motion equations and vector equations. Now, if the acceleration in the X axis is equal to zero, it actually simplifies a lot of our four equations here because the acceleration terms will go away. And these equations actually don't really are not really helpful to us in the third term, acceleration term goes away, Which means that the Onley equation that we're gonna use in the X axis for Project on Motion is just our equation for constant velocity. Now, in the Y axis where you have some acceleration that isn't equal to zero, it's actually equal to negative G. Then we're just gonna use our 3 to 4 equations of motion and that's really it. And the other thing is because we're moving from 1 to 1 dimension to two dimensional motion, we're all just gonna We're also just gonna need some general vectors. Equations? That's really it, guys, there's nothing new that we haven't already seen before how to dio. So let's go ahead and take a look at how to solve some problems.

2

Problem

ProblemWhich of the following quantities are constant during projectile motion?

A

Vertical acceleration & vertical velocity

B

Angle (direction) of the velocity vector

C

Horizontal acceleration & vertical velocity

D

Vertical acceleration & horizontal velocity

3

concept

## Solving Horizontal Launch Problems

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Hey, guys. So now that we've been introduced to the various kinds of project emotion, it's time to start solving some problems. So in this video, I'm gonna give you a great system for solving any one of your project, our emotions, problems that you might see. So we've got a list of steps and equations. Let's just get right to it. So we got a ball that rolls horizontally off of the table and we know the height and speed. We're gonna calculate this first part the time that it takes for the ball to hit the ground. So here's how these problems are always gonna go. Here's your first step. You're gonna draw the paths in the X and Y axis. So we got this ball that rolls horizontally off the table. First, we need to sketch out what the trajectory is. Once it leaves the table, it's gonna be under the influence of only gravity. It's gonna take a curve parabolic path here, so we're gonna draw the passing the X and Y axis. So what happens is, even though this is a two dimensional path from here to here, we could break this up and imagine if we could only move along the X axis that will be moving from here to here and in the Y axis. We're just moving from here down to here. So now the points of interest points of interest could be anything like your initial. So where your starting position is, that's always gonna be a and then your final. So the final is gonna be down here and then anything else that may happen in between, like you reaching a maximum height or something like that. Now, in this particular example were only told that we're going from the table down to the ground. So that's my initial final. So I'm just gonna use those two points of interest A and B. Now, there may be more later on, but just for now, we're just gonna use Thies too. So the second step is running, determined to target variable. In this case, we're looking for the time that it takes for the ball to hit the ground. So it's gonna be t then that brings us to the third step. Is which interval are you going to use? An interval is basically just between what? Two points in the diagram during the problem. Are you looking at? For example, when we go pick our equations here, if we're using equation like Delta X equals V X, this is a displacement. But between, what, two points. What interval are we looking at for that displacement? So in this particular problem, because we're looking for the time it takes for the ball to go from the table at point A to the ground at point B. Then the interval we're gonna use is the one from A to B. Okay, so now what equation do use to solve for T? And this is gonna be t a B now that we know the interval. Well, if you take a look at our equations, we have time in the X axis. The time also appears in the Y axis. In fact, in all of your project, on motion problems time, this variable can be found by either the X or the Y axis equations. So here's the deal. We have four equations in the y axis, but we only have one equation to use in the X axis. And because the X axis only has one equation, we're always gonna try the X axis first because it's the simplest thing to Dio. So we're gonna go ahead and do that in the X axis. We've got Delta X from A to B equals V X Times ta to be So this is our target variable. We just need to know the other two variables here. So what about the displacement? Well, the displacement is gonna be from point A to point B. So it's gonna be over here and this displacement here don't x from A to B. We don't know What about the X velocity? So we don't know what this is. What about this X velocity over here? Well, we have an initial velocity. We're told that the initial velocity of the ball is just 3 m per seconds. So in general, if we want the X component in the y component, we just have to use our vectors. Equations are cosine and sine functions. So what happens is my initial velocity in the X axis, which I'm gonna call the X because it's the ex velocity at point A. It's just gonna be v non times the cosine of theta. So it's gonna be three times the cosine of What angle do I use now? Well, at this velocity is purely horizontally along the X axis like this. Then that means that the angle this makes with the X axis is just zero. So we're gonna use three times the coastline of zero and that's three in the y axis we're gonna do. The same thing is V A y. And this is gonna be even on time to the sign of data. So it's gonna be three times the sign of zero in a sign of zero is just zero. So this ends of just being zero here. So we have this initial velocity in the X axis that this still isn't enough, right? We still only have two. We only just have one variable. We have two unknowns. So this is a situation that may happen often in your physics problems. And if you ever get stuck and you can't solve using an x x x x axis equation that you could always try to solve it with a Y axis equation, you could go to the other access and try to solve for their For instance, we got stuck here in the X axis, so we're now we're gonna go ahead and try to solve in the Y axis, so we're gonna pick one of these equations to use. But in order to do that, we need three out of five variables. So here we're gonna list out all our variables here. The first one is the easiest. This is just the acceleration, which is always negative. 9.8. Now you've got the initial velocity. Remember that This the initial y velocity at point A. We know that's equal to zero. What about the final velocity? That's the final velocity when it gets to point B, and we actually don't know what that is, that would be the Y velocity here at the ground, which we don't know, and then we've got dealt it. Why? From a to B, that's the vertical displacement. And then we've got time from A to B, and that's actually we're looking for here. Okay, so we don't know what the why velocity is. What about this vertical displacement? Well, the vertical displacement from A to B is just gonna be the distance vertically that this covers from the table down to the ground. We're told that the ball or that the table's height is 2 m, so basically the ball is going to fall 2 m to the ground. However, so this is gonna be our delta y. So because this displacement is actually gonna be down boards where energy, We're gonna end up at a lower height than we started from then. This is actually gonna be a negative 2 m, because, remember, we're always gonna use the convention that up into the right is positive, so any downward displacements are gonna be negative. So we've got that Delta Y and we know that's negative to. And now we have our three out of five variables. We've got 12 and three. So we're gonna pick the equation that does not have the final velocity. And if you look, there are equations here, that's gonna be equation number three. So we're gonna use equation number three, and they were gonna use all of our subscript So Delta y from A to B is gonna be the initial velocity, which is va y times t a B plus one half a y t a b squared. So remember that this v a y just turns out to be zero So now we can simplify and then we just solve for t A B. So this delta Y is negative two. Then we've got one half times negative, 9.8 times ta b squared. And now we just solved. So if you move all of these terms over to the to the left side, it's becomes two times negative. Two divided by negative. 9.8. That's gonna be ta b squared. And so are ta be ends up being when you take the square root 0.64 seconds. So that's the answer to party. Okay, so let's move on now to part B. So we're gonna go through the same list of steps again. So in part B, we're looking for the horizontal displacement off the ball so we don't have to do first. The first step again. We've already had the path and the X and Y what about this target variable? What are we looking for here? Well, this is a displacement, so it's gonna be either Delta X or Delta Y um, but because we're looking for a horizontal displacement, this is gonna be Delta X. That's the second step. The third step is between What interval? We already know we're working from the interval from A to B. So this is gonna be Delta X from A to B here. So now we're just gonna go and to the, um, equations. Right? So which equation do we use to solve? Well, this is an X axis variable Delta X. So we're gonna use the Onley X axis equation. So this is gonna be Delta X from A to B equals V X Times ta to be is actually just the same exact equation that we started off with in the first part. But now we actually know what this time is, so we can actually go ahead and solve because we have both of these variables here. So Delta X from A to B. It's just my 3 m per second times, the 0.64 and that's gonna be 1.92 m. And so that is the answer to the second part. All right, so that's how we solve any one of your project on motion problems. So I wanna make some points here because the type of problem that we just solved was called a horizontal launch So this happens whenever you launch an object horizontally and special points. You should know about this. The first is that the initial velocity is Onley in the X axis. What we saw is that when you launch something purely horizontally, all of its velocity is in the X axis. So your V not X Your initial velocity is just the V not that you were given And remember all objects in project the emotion have zero acceleration in the X axis. So what that means is that your V X will never change Whatever this initial velocity is in, the X axis will always be the same exact number throughout the whole entire motion. And the last thing is that because an object has launched horizontally than the initial component of its velocity in the Y, Axis V not why is just equal to zero. All right, so that's how to solve these kinds of problems. Let me know if you guys have any question

4

Problem

ProblemA rock is thrown horizontally with a speed of 20 m/s from the edge of a high cliff. It lands 80 m from the cliff's base. How tall is the cliff?

A

78.4 m

B

19.6 m

C

122.5 m

D

24.5 m

5

example

## Clearing the Net

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Hey, guys. So this problem We were trying to figure out the initial speed that a Ping Pong ball needs to have in order to clear a net on the table. We know that this Ping pong ball is served horizontally, so let's go ahead and just draw a quick diagram of what's going on. So we have a ball that's gonna be over here, it's gonna be served perfectly horizontally. We know this ball is served 1.2 m off of the ground, and what it needs to do is some distance later. It needs to clear some nets. So I'm gonna draw this little point right here. So what happens is this ball needs to have some initial velocity so that when it follows its parabolic path, it's gonna just barely go over this net right here. So that's what this sort of path looks like right here. We're told also that this net is 1.6 m away from the player. So this is we know this distance here is 1.6 and we know this heights here of the net off of the table is 0.9. Alright, so that's what we know about this problem. So now we're just gonna go ahead and draw the past in the X and Y We're just gonna stick to the steps, right? So we're gonna draw the paths in the X and Y axis. So this is my ex access. It just goes from straight from here to here and the Y axis. If it served horizontally, then in the Y axis is just gonna go from here all the way down to this point over here. So what are points of interest? We have our initial, which is point a, and then nothing else happens in between. But basically, it's gonna hit the Nets, or it's gonna go right over the nets. So that's gonna be our final point. Point B. So those are our paths in the X and Y. We've got from A to B. And so now we're going to figure out what target variable we're looking for. What is that initial speed? What we actually looking for here? What? We're looking for V knots, right? So this needs to have some initial velocity V knots. However, what we do know from initial velocities is that we do know about horizontal launches is that the initial velocity is purely in the X axis. So we're actually looking for is V not or Vienna X, But this is also just the same thing as Vieques. It's basically just the same horizontal velocity throughout the entire motion. So all these three variables really mean the same thing? So that's really what we're gonna be looking for. So now that brings us to the third step. What interval or we're gonna look at We're looking for the initial velocity and we're looking basically just between the time that it served and the point where it hits the net or goes over the net, Then we're just gonna look at the interval from A to B. Okay, so we're looking for a X variable The velocity in the X axis We're just going to start off with the X axis equation. There's only one which says that Delta X from A to B is equal to v x times t from A to B. So this is what I'm looking for. I'm looking for the initial velocity in the X axis, so I'm gonna need the horizontal displacement and I'm gonna need the time So let's take a look at the horizontal displacement. We were told that this Ping pong ball is served 1.6 m away from the net. So that means that this is basically my Delta X. So this is my delta X from A to B, and this is equal to 1.6. So I'm just actually gonna erase this over here because it's actually Mawr. Ah, sort of. You could see it better here that it's 1.6. All right, so we know what this number is. Unfortunately, we don't know how long it takes to go from A to B, so we actually don't know how long that is. So we're a little bit stuck with this equation. We can't use it. So therefore, when wherever we're stuck in the x axis, we're gonna have to go to the Y axis. So we go over here to look for the y axis because I want time. So I'm gonna listen at all my variables. A y equals 9.8 v. Not y is v a Y v y is v b y. We've got Delta y from A to B and we've got time from a to B. Remember, we're looking for the time here, so we can actually plug it back into this equation over here. So we're looking for the time. That's our target variable. So we got we need three out of five variables. So what about the initial velocity in the Y Axis V A. Y? Well, remember, what's special about horizontal launch problems is that your V a y. Your initial velocity in the Y axis is just equal to zero. And so all of your initial velocity is just purely in the X axis. And that's just the speed that was given to us. Or actually, in this case, the speed that we're gonna look for this is V X. All right, so we know that the initial velocity in the Y Axis zero What about the final velocity here at point B? So what is the components of the velocity V B Y? We actually don't know what that is. So this is gonna be We don't know what that is. What about the delta y from A to B? What about the horizontal? Sorry, The vertical displacement from A to B Well, from from A to B. We're actually falling some distance here, but what number we're gonna use are we gonna use the 1.2 or we're gonna use the 0.9. Well, it's actually just gonna be the difference between those two numbers, because in our path, from A to B or actually just falling this distance over here, which, if you think about it, is gonna be the 1.2 m that we're serving off of the ground to 0. m that the net is above the ground. So it's whatever the difference of those two numbers is. So it means that Delta Y from A to B is actually negative 0.3. It's negative because this displacement points downwards. So this is gonna be negative 0.3, and so that gives us three out of five variables. So this one is the ignored, and we pick our equation based on the ignored variables. That's actually gonna be equation number three, the one that ignores the final velocity. So Delta y from A to B is going to be, um, let's see. I've got v a Y t a b plus one half a Y t a B squared. Alright, so it's again. It's also useful about horizontal launch equations. Or what's convenient is that this term will always go away. Because remember that V A Y is equal to zero. So the only just leaves us with two terms. So now I could just go ahead and fill in everything. I know what Delta Y is. It's just negative. 0.3, this is gonna be negative. One half negative, 9.8, that's G. And then you got ta be squared. So if you go ahead and soul for this, what you're gonna get is the TA be is just equal to 0. seconds. So now we're just gonna plug this back into our X axis equation and then we'll be able to figure out that horizontal velocity So my delta X from A to B equals v x times ta to be Now I have both with those numbers are and so the final velocity or the initial velocity is going to be the Delta X, which is the 1.6 divided by 0.25. I get 6.4 m per second and that is the answer this Ping pong ball needs to hit. Needs to be hit at 6.4 m per second in order to clear the net on the other side. That's it. But this one, guys, that actually is answer choice D. So let me know if you guys have any questions.

6

example

## An Unfortunate Landing

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Hey, guys, let's take a look at this problem here. You're gonna take a ball horizontally off of the building. And what happens is a car on underneath you or below on the street is gonna accelerate, and then your ball is gonna land on the car. We're gonna figure out the acceleration of the car is before we go into anything. Let's just go ahead and draw this out because this looks a little bit different than some of the problems we've seen before. So we've got the roof of the tall building like this. So I've got a ball that's on top here. I'm gonna kick it with some horizontal velocity. And so therefore, it's gonna take this sideways or parabolic path as it goes down. Now, at the same time, there's a car that's at the bottom right on on the street, right underneath you. So let's say this is a little car like this, and it's gonna accelerate from rest, and eventually what happens is that their ball while it's in flights, the car is gonna be moving, and eventually the car is gonna land underneath the ball. Basically, they're just gonna catch up to each other. So this is a little bit different than the kinds of problems that we've seen so far. So we want to do is we want to figure what the acceleration of the car is. So this is just a sorry, not zero. We're actually looking for what this variable is. So because this is ah type of problem in which one object is going to catch up to another one while it's falling, then we're gonna do is we're gonna follow the steps that we use to solve catch problems. And remember that from catch problems, we have a Siris of steps. First, we're gonna right the position equations for both of them. Then what we do is set them equal to each other so set equal. And the third step is we solve for time. Now, we're basically gonna do those exact steps over here. Let's just go ahead and start out with the position equations for both of these objects. So, first, before we do that, we're gonna draw the paths in the x and Y axis and step one of our project on motion problems. We know that if this object is gonna take this parabolic path like this. We can split up the X motions in the UAE emotions. So it looks like this. So that's the path of the X and Y. What of our points of interest? Well, really, we're just kicked from the roofs. That's point a, That's our initial. And then nothing happens in between. And then it finally just lands on the car on the ground at point B. So those are passed in the X and Y axes. So what is our position equation for the ball Look like? What happens is that we sort of just imagine that this wall here is X initial and we call that zero, then the X of your ball is just gonna be x knots, plus the velocity of the ball Times t plus one half of the acceleration in the X axis times time squared and the ex of your car is gonna be x non plus the car initial T plus one half a car Times T squared. So I'm gonna call the acceleration of a car, a car, and that's actually what we're gonna be looking for here. That's the target variable. So we're looking for what is a car. Okay, so now that we've written out the position equations, let's see if we can simplify them. Well, if we just call the initial position zero, then both of these terms will go away. Now, what about the ball? The ball has an initial velocity, so we won't get rid of that term. But remember that in the X axis, the acceleration is equal to zero. And the only acceleration for the ball happens in the Y axis because of gravity that x downwards. But in the X axis acceleration zero. So that term also goes away, and then for the car. What happens is we're told that the car accelerates from rest, which means that the V initial of the car is gonna be zero. So really, what these equations actually simplify Teoh is we just have V ball times time and then this is just gonna equal to one half a car. Times t squared. So really, it's just those two equations here, So that's step one. The second thing we're gonna do now is we're actually just going to set them equal to each other. So we have to just set X ball equal to x car. So that means that we're gonna do is we're gonna have the ball times t equals one half of a car. Times t squared, OK? And so now the third step is we're just gonna solve for the time. So notice here how we can actually cancel out a factor of time for both sides of the equation here, and then we end up with is we just end up with the ball equals one half of a car times one factor of t remember, in this equation or in this problem, we're looking for a car. Now what is V Ball? Well, the ball is just the initial velocity that's given to the ball. And we know that V ball, which is really just the velocity in the X axis is just eight. We also know that because this ball is launched purely horizontally, that v a y is just gonna be zero. So that's gonna be, um, so remember. That's always gonna be useful in your horizontal launch problems. So we do know what the ball is. Unfortunately, if we're trying to figure out the acceleration, I don't know what the time is. That's the time that it takes for the ball toe, actually travel through the air and then finally hit the ground. And so, just like any projectile motion problem here, I've gotten stuck in the X axis trying to solve for time. So what do I do? I just go to the Y axis and the Y axis. There's only one thing that's moving. It's just the ball that's going from the roof down to the ground. So all of these y axis variables, they're gonna be for the ball. The acceleration is 9.8, the initial velocity in the Y axis va y Because we just said zero. What about the final velocity? That's gonna be the velocity at point B. We've got Delta Y A B. And then ta be Remember, I came to the Y axis because I'm looking for the time. If I could do that, if I could figure that out, you can plug it back into this equation and solve the acceleration. So I need one more variable here. So the final velocity in the Y axis would just be V B y like this. I don't know what that is. What about the vertical displacement? What about Delta y from A to B. Well, if I look at my diagram here and if I look at my problem, what I'm told is that the ball is kicked at 8 m per second from a 40 m tall building. So what happens is when it falls from a to B. This vertical displacement here is 40 m. However, because this vertical displacement is downwards, then this is just gonna be a negative 40 m. So we know this Delta Y A b is negative 40. And so now we have 3 to 5. So three out of five and we picked the equation. Whoops. And we're going to pick the equation that ignores my final velocity. So if you look through your equations here, that's going to be equation number three. So this is gonna be Delta y A P equals and this is gonna be v A y Times t A B plus one half a y t a b squared, remember. And horizontal launch problems. Um, your V A Y is equal to zero. And so this term always goes away. So you just plug in everything you're gonna have a negative 40 equals one half negative, 9.8 times ta b squared. And if you go ahead and solve for this, what you're gonna get is when you rearrange everything, you're gonna get two times negative, 40 divided by negative, 9.8 and then you're gonna square root that number that's gonna be ta be, and you're gonna get to 0.86 seconds. So now that we finally have this time here, we can plug it back into our equation and solve for the acceleration. So we've got the ball, which is equal to eight. That's gonna equal one half times a car, which is what we're looking for times to 86 seconds and you go ahead and work this out. Where you gonna get Is that a car? The acceleration of the car is equal to 5.60 m per second squared. So that's how fast the car has to accelerate in order to basically fall right underneath the ball as it hits the ground. So that is answer choice. Be so hopefully you guys can see you have to be on the lookout for these kinds of hybrid problems where you really just combining multiple kinds of problems like a catch problem with also a project on motion. But if you just follow the steps for both of them, you'll still get the right answer. That's about this one, guys, let me know if you have any questions.

Additional resources for Intro to Projectile Motion: Horizontal Launch

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