Hey, guys. So previously, what we've seen in our projectile motion problems is that we'll get stuck in one axis. We'll have to go to the other axis to get that variable and then bring it back to the original equation. I'm just going to fly through this example really quickly. It's an example that we've already seen before, this previous example here. Let's check it out. We've got a horizontal launch. We're trying to figure out the delta x from a to b. So we start off with our delta x equation and we have the x velocity, but we don't have the time. And whenever we get stuck in one axis, we're just going to go to the y axis for instance, and then we're going to figure out our variables and we're going to try to solve for this t. So we set up an equation to solve for time. We get a number and then we basically just plug it all the way back into our original expression and then we can figure out delta x. This is just equal to 3 times 0.64, and then we just get a number 1.92 meters. We've already seen this before, very straightforward. Every time we get stuck, we just go to the other axis.

What I want to show you in this video is that there may be some situations in which you actually get stuck in both the x and the y axis when you try to go to the other one. So to get out of this situation, we're going to use a method called equation substitution. This usually happens whenever you're solving problems and you have two out of these three variables that are unknown, the initial velocity, the angle, and the times. They may be unknown, but they may not necessarily be asked for. Let me just go ahead and show you how this works using this problem here. So we've got a soccer ball that's kicked upwards from a hill, and it lands, you know, it's in the air for 4 and a half seconds and lands 45 meters away. And we're going to figure out the initial velocity and angle at which the soccer ball is kicked. So let's go ahead and draw a diagram. Right. So we've got this hill like this and we've got a soccer ball that's kicked at some angle. So it's going to go like this. It's going to be like an upward launch and then it's going to land down here. Right. So we're going to go ahead and just go through the steps, draw the path in the x and y. Let me just scoot this down a little bit. Give me some room. So the x axis would look like this and the y axis, this would look like this. And our points of interest are a, the maximum height, b, the point where it returns to the original height, c, and then back down to the ground again. That's d. So in the x axis, this would look like this, a, b, c and then finally d. In the y axis, this would go up and then back down through c and then down to d again. We've already seen the situation before. Cool. So what's the next step? So now we just need to figure out the target variables. Well, there are two of them. In this case, we're looking for the initial velocity and the angle, and so those are going to be our unknown variables. So what interval are we going to use? Well, if you take a look at this problem here, the thing we know the most amount of information about is the interval from a to d. The one thing we know is that in the x axis, the total amount of time that it spends in the air, t from a to d, is 4.5 seconds. And we also know that delta x, the total horizontal displacement, is 45 meters. That's where the ball lands 45 meters away. So we're just going to use the interval from a to d, because that's the one we know the most information about. So let's start off with the x axis. Our only equation that we can use is delta x from a to d is equal to vxa-d⋅Ta-d. So we actually know both of these equations over here. So we have, delta x is 45. And what about this initial velocity here? Well, the initial velocity is going to be in the x axis, but that's not what we're solving for. We're trying to solve for v₀ and θ. And the way that we would normally solve this is by using vector decomposition. If we want vx, we would use v₀⋅cos(θ). So I'm going to rewrite this vxa-d term in terms of v₀ and θ. So basically, what this turns into is I have 45 equals v₀⋅cos(θ)⋅4.5. And if I go ahead and divide this to the other side, I get 10 equals v₀⋅cos(θ). Now, unfortunately, there's a problem here as I still have two unknowns in this problem. My v₀ and my θ are both unknown. So I'm stuck here. I can't solve for either of them. So what do I do? Whenever I'm stuck, I'm just going to go to the y axis and then try to solve for those variables there. I'm going to use the same exact interval, the interval from a to d, but now I need my 3 out of 5 variables. I've got 9.8, is my, sorry, negative 9.8 is my aya-d. Now the initial velocity is going to be my vaya-d. Final velocity is going to be vdya-d, and then I've got delta y from a to d, and then I've got t from a to d. Well, t from a to d I already know is 4.5. What about delta y? Well, that's just the vertical displacement from a down to d. I know that this is a 5 meter hill so it's going to be negative because I'm going downwards and that's my delta y. So and this is negative 5. And then what about my vdya-d? I don't know what the final velocity is in the y axis. What about vaya-d? Well, the way that I would normally solve this is just like vx is v₀⋅cos(θ), my v₀y}, which is vaya-d, is just equal to v₀}⋅sin(θ). So I'm just going to plug in this expression here for vaya-d. This is just equal to v₀}⋅sin(θ). Now if I were to set up an equation for this, this would be my ignored variable, and even though I don't know v₀} or sin(θ), if I wanted to find it, I have my 3 out of 5 variables, so I'm just going to use equation number 3, which says that delta y from a to d is equal to vaya-d⋅Ta-d+12⋅ay×Ta-d2. So if I go ahead and plug in everything I know about this problem and then I also replace my vaya-d with v₀}⋅sin(θ), then what I'm going to get is I'm going to get negative 5 equals