Intro to Projectile Motion: Horizontal Launch - Online Tutor, Practice Problems & Exam Prep

Introduction to Projectile Motion

Hey, guys. For the next couple of videos, we're already 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 in projectile motion and it might seem that there are different ways to solve them. So what I'm going to show you in this video is I'm going to introduce the main types of projectile motion that you'll see, and I'm going to show you that to solve them, we're just going to use equations that we've already seen before. So let's check it out.

Projectile motion is going to happen whenever you take an object and you throw it, and it moves in 2 dimensions, that's the important part, under the influence of only gravity. So we've actually seen something similar to this before when we studied 1-dimensional vertical motion. If you take an object and you throw it straight up, then it comes straight back down along the same direction that you threw it, because it's under the influence of only gravity. It just moves purely up and down.

What's different about projectile motion is that we're not going to throw something up and down. We're going to give it some sideways velocity, whether we're throwing it perfectly horizontal, downwards, or upwards. These are really the main types that you might see. But because we're giving it some 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 2-dimensional parabolic paths. Some other examples that you might see are where you throw something downwards or even upwards. So you might see situations where it returns to the same height, or it 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 going to fall into one of these categories.

So what'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 2-dimensional parabolic paths. So how do we solve problems? Well, remember, whenever we have physics problems in 2 dimensions, whether it's motion or vectors, we can always decompose them into 1-dimensional x and y parts, and it's no different for projectile motion.

So what 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 only 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 going to pick one over here. So like for instance, from a, and then it goes up to some point. Let's call that B. Then it goes back down again. Let's call that point C. We can just take this and break it up into its two one-dimensional parts in the X and Y. So if you only could move along the X-axis, then in the X-axis, you would just go from a to b to c like this.

In the Y-axis, if you could only move along the Y-axis, what this projectile would look like is it would go from a and then it would go up to b, and then it would go back down to c. So it would actually look a little bit like this motion over here. So really, it's just these two one-dimensional motions that are happening at the same time. Now why is this important? Because we take a look here, the only acceleration that this object experiences is just going to be in the Y-axis due to gravity.

So when it comes to these two one-dimensional motions, projectile motion is really just a combination of horizontal motion where the acceleration in the x-axis is equal to 0 and vertical motion where the acceleration in the Y-axis is equal to negative g. It's negative because we're always going to assume that up into the right is positive, and g acts downward, so it's negative.

Now why is this important? Because when it comes to the equations that we use for projectile motion, we're really just going to be using combinations of motion equations and vector equations. Now if the acceleration in the x-axis is equal to 0, that actually simplifies a lot of our four equations here because the acceleration terms will go away, and these equations actually aren't really helpful to us. In the third term, the acceleration term goes away, which means that the only equation that we're going to use in the x-axis for projectile motion is just our equation for constant velocity.

Now in the Y-axis, where you have some acceleration that isn't equal to 0, and it's actually equal to negative g, then we're just going to use our three to four equations of motion, and that's really it. And the other thing is because we're moving from one to one dimension to 2-dimensional motion, we're also just going to need some general vectors equations. That's really it, guys. There's nothing new that we haven't already seen before how to do. So let's go ahead and take a look at how to solve some problems.

Which 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

Solving Horizontal Launch Problems

Hey, guys. So now we've been introduced to the various kinds of projectile motion, it's time to start solving some problems. In this video, I'm going to give you a great system for solving any one of your projectile motion problems that you might see. We've got a list of steps and equations. Let's just get right to it.

We've got a ball that rolls horizontally off of a table and we know the height and the speed. We're going to calculate this first part, the time it takes for the ball to hit the ground. Here's how these problems are always going to go. Your first step is to draw the paths in the x and y axes. We've 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 going to be under the influence of only gravity. It's going to take a curved parabolic path. We're going to draw the path in the x and y axes.

What happens is even though this is a two-dimensional path from here to here, we can break this up and imagine if we could only move along the x-axis that we'd be moving from here to here, and in the y-axis, we've just been 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 going to be point A. And then your final, so the final is going to be down here and then anything else that may happen in between like you reaching a maximum height or something like that. In this particular example, we're only told that we're going from the table down to the ground. So that's my initial and final. So I'm just going to use those two points of interest, A and B.

The second step is we need to determine the target variable. In this case, we're looking for the time it takes for the ball to hit the ground. So that's going to be t. The third step is which interval are we going to use? An interval is basically just between what two points in the diagram or in the problem are you looking at. For example, when we go pick our equations here, if we're using an equation like Δx=v_xt, this is a displacement. But between what two points, what interval are we looking at for that displacement?

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, the interval we're going to use is the one from A to B. Now what equation do we use to solve for tAB now that we know the interval? Well, if you take a look at our equations, we have time in the x-axis, but time also appears in the y-axis. In fact, in all your projectile motion problems, time, this variable, can be found by either the x or y-axis equations. 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 going to try the x-axis first because it's the simplest thing to do. In the x-axis, we've got Δx=v_xt from A to B. So this is our target variable. We just need to know the other two variables here.

What about the displacement? The displacement is going to be from point A to point B. What about the x velocity? Well, we have an initial velocity. We're told that the initial velocity of the ball is just 3 meters per second. So in general, if we want the x component and the y component, we just have to use our vector's equations, our cosine and sine functions. So my initial velocity in the x-axis, which I'm going to call v_ax because it's the x velocity at point A, is just going to be v₀·cos⁡(θ), and it's going to be 3·cos⁡(0) which is 3. In the y-axis, we're going to do the same thing v_ay=v₀·sin⁡(θ), so it's going to be 3·sin⁡(0), and the sine of 0 is just 0. So this ends up just being 0 here. We have this initial velocity in the x-axis, but this still isn't enough. We still only have one variable. We have two unknowns.

This is a situation that may happen often in your physics problems. If you ever get stuck and can't solve using an x-axis equation, then you could always try to solve it with a y-axis equation. You can go to the other axis and try to solve for it there. For instance, we got stuck here in the x-axis. So now we're going to go ahead and try to solve in the y-axis. We're going to pick one of these equations to use, but to do that we need 3 out of 5 variables. We've got the acceleration which is always -9.8. We've got the initial velocity. Remember that this is the initial y velocity at point A, and we know that that's equal to 0. We don't know the final velocity, which would be the y velocity here at the ground.

What about this vertical displacement? The vertical displacement from A to B is just going to be the distance this ball covers from the table down to the ground. We're told that the table's height is 2 meters. So basically, the ball is going to fall 2 meters to the ground. So, this is going to be our Δy, and because this displacement is actually going to be downwards, we're ending up at a lower height than we started from, then this is actually going to be a negative 2 meters because we're always going to use the convention that up and to the right is positive. So any downward displacements are going to be negative.

We've got that Δy and we know that's negative 2, and now we have our 3 out of 5 variables. So we're going to pick the equation that does not have the final velocity. Equation number 3, so we're going to use equation number 3, and we're going to use all of our subscripts. Δy=v_ayt+12a_yt2. This v_ay turns out to be 0. So now we can simplify and solve for tAB. This Δy is negative 2, and we've got 12·-9.8 times tAB·2. We now solve this equation and tAB ends up being the square root, which is 0.64 seconds. That's the answer to part A.

Let's move on now to part B. We're going to go through the same list of steps again. In part B, we're looking for the horizontal displacement of the ball. We don't have to do the first step again, we already have the path in the x and y. What about this target variable? We're looking for horizontal displacement, this is going to be Δx. The third step is between what interval? We already know we're working from the interval from A to B. So this is going to be Δx+v_xtAB here. So now we're just going to go to the UAM equations. Right? Which equation do we use to solve? Well, this is an x-axis variable Δx. So we're going to use the only x-axis equation.

This is going to be Δx=v_x·tAB, and this 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 Δx from A to B is just my 3 meters per second times the 0.64, and that's going to be 1.92 meters. That is the answer to the second part.

Alright. So that's how you solve any one of your projectile motion problems. So I want to make some points here because the type of problem that we just solved was called a horizontal launch. This happens whenever you launch an object horizontally, and there are some special points you should know about this. The first is that the initial velocity is only in the x-axis. What we saw is that when you launch something purely horizontally, all of its velocity is in the x-axis. Your v_0x, your initial velocity is just the v₀ that you were given. Remember, all objects in projectile motion 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 is launched horizontally, then the initial component of its velocity in the y-axis v_0y is just equal to 0.

Alright? So that's how to solve these kinds of problems. Let me know if you guys have any questions.

A 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

Clearing the Net

Hey, guys. So in this problem, we're 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 going to be over here. It's going to be served perfectly horizontally. Clear some nets. So I'm going to 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 going to 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 meters away from the player. So we know this distance here is 1.6. And we know that this height 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 going to go ahead and draw the path in the x and y axis. This is my x-axis. It just goes straight from here to here. And the y-axis, if it serves horizontally, then in the y-axis, it's just going to go from here all the way down to this point over here. So what are our points of interest? We have our initial, which is point A, and then nothing else happens, in between, but basically, it's going to hit the net or it's going to go right over the net. So that's going to be our final point, point B. So those are our paths in the x and y. We've got from A to B. So now we're going to figure out what target variable we're looking for. What is that initial speed? What are we actually looking for here? We're looking for v₀ (v naught). However, what we do know about horizontal launches is that the initial velocity is purely in the x-axis. So what we're actually looking for is v₀, or v_0x, but this is also just the same thing as v_x. 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 going to be looking for. So now that brings us to the 3rd step. If we're looking for an x variable, the velocity on the x-axis, we're just going to start off with the x-axis equation. There's only one which says that Δx from A to B is equal to v_x × t from A to B.

So I'm looking for the initial velocity on the x-axis. So I'm going to need the horizontal displacement, and then I'm going to need the time. So let's take a look at the horizontal displacement. We're told that this ping pong ball is served 1.6 meters away from the net. So that means that this is basically my Δx. So this is my Δx from A to B, and this is equal to 1.6. 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 we're stuck on the x-axis, we're going to have to go to the y-axis. So let's go over here to look for the y axis because I want the time. So we list out all our variables: a_y = 9.8, v_0y = 0 (because the initial velocity in the y-axis is 0 for horizontal launches), v_y is unknown, Δy from A to B = (0.9 - 1.6) = -0.7 meters (falling downwards), and time from A to B. Remember, we're looking for the time here so that we can plug it back into this equation over here. We're looking for the time. That's our target variable.

This is v_0x. We know that the initial velocity in the y-axis is 0. What about the final velocity here at point B? We don't know that. What about the vertical displacement from A to B? Well, from A to B, we're actually falling some distance here. But what number are we going to use? We're going to use the 1.6 or the 0.9? Well, it's actually just going to be the difference between those two numbers. Because in our path from A to B, we're actually just falling this distance over here, which if you think about it, is going to be the 0.9 meters that the net is above the ground. So, delta y from A to B is actually negative -0.7. So that gives us 3 out of 5 variables. So this one is ignored, and we pick our equation based on the ignored variables.

So that's actually going to be equation number 3, the one that ignores the final velocity. Delta y from A to B is going to be vaytab+12ayt2. So, now I can just go ahead and fill in everything. I know what delta y is. It's just -0.7. This is going to be -1/2 9.8 and then we've got t_AB². So if you go ahead and solve for this, what you're going to get is that t_AB is just equal to 0.25 seconds. So now we just 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 = v_x × t from A to B. Now I have both of what those numbers are. And so the final velocity or the initial velocity is going to be the delta x, which is 1.6 divided by 0.25, and I get 6.4 meters per second. And that is the answer. This ping pong ball needs to hit needs to be hit at 6.4 meters per second in order to clear the net on the other side. That's it for this one, guys. That is actually answer choice D. So let me know if you guys have any questions.

An Unfortunate Landing

Hey, guys. Let's take a look at this problem here. You're going to kick a ball horizontally off of a building. What happens is a car below on the street is going to accelerate, and then your ball is going to land on the car. We're going to figure out what the acceleration of the car is. So before we go into anything, let's just go ahead and draw this out because this looks a little different than some of the problems we've seen before. We've got the roof of the tall building like this. I've got a ball that's on top here, I'm going to kick it with some horizontal velocity. And therefore, it's going to 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 the street right underneath you. Let's say this is a little car like this, and it's going to accelerate from rest and eventually what happens is that their ball, while it's in flight, the car is going to be moving and eventually the car is going to land underneath the ball. Basically, they're just going to catch up to each other. This is a little different than the kinds of problems that we've seen so far. So, what we want to do is we want to figure out what the acceleration of the car is. This is just a, whoops, sorry, not 0. We're actually looking for what this variable is. Since this is a type of problem in which one object is going to catch up to another one while it's falling, then we're going to follow the steps that we use to solve catch problems. Remember that from catch problems, we have a series of steps. First, we're going to write the position equations for both of them. Then what we do is set them equal to each other. Lastly, the third step is we solve for time.

We'll start with the position equations for both of these objects. Before we do that, we're going to draw the paths on the x and y axes in step 1 of our projectile motion problems. Knowing that if this object is going to take this parabolic path like this, we can split up the x motions and the y motions, so it looks like this. What are our points of interest? Really, we're just kicked from the roof so that's point A, it's our initial. Then nothing happens in between and then it finally just lands on the car on the ground at point B. So those are our paths on the X and Y axes. What does our position equation for the ball look like? If we imagine that this wall here is x initial and we call that 0, then the x of your ball is just going to be x₀ + velocity of the ball times t + one-half of the acceleration on the x-axis times time squared. And the x of your car is going to be x₀ + v_car initial t + one-half a_car times t squared. I'm going to call the acceleration of the car a_car, and that's what we're looking for here. That's the target variable.

Now that we've written out the position equations, let's see if we can simplify them. If we call the initial position 0, then both of these terms will go away. Now what about the ball? The ball has some initial velocity, so we won't get rid of that term. But remember that on the x-axis, the acceleration is equal to 0. The only acceleration for the ball happens on the y-axis because of gravity acting downward. But on the x-axis, the acceleration is 0, 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 going to be 0. So really, these equations actually simplify to just v_ball times time, and this is going to equal 1/2 a_car times t squared. So it's just those two equations here.

Next, we're just going to set them equal to each other. We have to set x_ball equal to x_car. So what that means is we're going to have v_ball times t equals one-half of a_car times t squared. Now the third step is we're just going to solve for the time. Notice here how we can cancel out a factor of time from both sides of the equation. What we end up with is v_ball = 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, v_ball is just the initial velocity given to the ball, which is really just the velocity in the x-axis, is just 8 m/s.

If we're trying to figure out the acceleration, and I don't know the time, it's the time that it takes for the ball to actually travel through the air and then finally hit the ground. Just like any projectile motion problem, I've gotten stuck on the x-axis trying to solve for time. So what do I do? I go to the y-axis. In the y-axis, there's only one thing moving. It's just the ball going from the roof down to the ground. So all of these y-axis variables are for the ball. The acceleration is 9.8 m/s². The initial velocity in the y-axis, v_a y, is 0. What about the final velocity? That's going to be the velocity at point B. We've got delta y from a to b and then t from a to b. I'm looking for the time. We know the vertical displacement from a to b, which when the ball is kicked at 8 m/s from a 40-meter tall building, this vertical displacement here is 40 meters. However, because this vertical displacement is downwards, then this is just going to be -40 meters.

We now have 3 out of 5 variables, and we pick the equation that ignores my final velocity. Delta y_a_b equals v_a_y times t_a_b plus one-half a_y t_a_b squared. In horizontal launch problems, your v_a_y is equal to 0, and so this term always goes away. Plugging in everything, -40 equals 1/2 (-9.8) times t_a_b². Solving for this, you get 2 * (-40) / (-9.8), and then square rooting that number, you get t_a_b and you get 2.86 seconds. Now that we finally have this time, we can plug it back into our equation and solve for the acceleration. v_ball, which is 8, is going to equal 1/2 times a_car, which is what we're looking for, times 2.86 seconds. Working this out, you get that a_car, the acceleration of the car, is 5.60 m/s². That's how fast the car has to accelerate in order to land right underneath the ball as it hits the ground. So that is answer choice B. So hopefully, you guys can see that you have to be on the lookout for these kinds of hybrid problems where you're really just combining multiple kinds of problems like a catch problem with also a projectile motion. But if you just follow the steps for both of them, you'll still get the right answer. That's it for this one, guys. Let me know if you have any questions.