1

concept

## Solving Projectile Motion Using Equation Substitution

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Hey, guys. So previously what we've seen in our project on motion problems is that we'll get stuck in one access. We'll have to go to the other access to get that variable and then bring it back to the original equation. I'm just gonna fly through this example really quickly. It's an example 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 ex equation and we have is we have the X velocity, but we don't have the time. And whenever we get stuck in one axis, we're just gonna go to the Y axis, for instance, and then we're gonna figure out our variables we're gonna try to solve for this tea. So we set up in 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 three times 0 64 and then we just get a number 1.92 m So 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 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 gonna use a method called equation substitution. So this usually happens whenever you're solving problems and you have to 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. You go ahead and show you how this works, using this problem here. So we got a soccer ball that's kicked upwards from a hill, aunt. It lands, you know, it's in the air for 4.5 seconds and lands 45 m away, and we're gonna figure out the initial velocity and angle at which the soccer ball was 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 gonna go like this. It's gonna be like, uh, upward launch and then it's gonna land down here, right? So we're gonna go ahead and just go through the steps, draw the past in the X and Y Let me just scoot this down a little bit, give myself some room. So the x axis that was to look like this and the y axis is would look like this on our points of interest are a the maximum height. Be point where returns to the original height C and then back down to the ground again. That's d so in the X axis is would look like this ABC and then finally d in the Y Axis is would go up and then back down through C and then down to Deacon. We've already seen the situation before. Cool. So what's the next step? So now we just need to figure the target variables. Well, there's two of them. In this case. We're looking for the initial velocity and the angle on DSO those gonna be are unknown variables. So what interval we're gonna 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 a. D is 4.5 seconds and we also know that Delta X, the total horizontal displacement is 45 m. That's where the ball lands 45 m away. So we're just gonna 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. Are only equation that we can use is Delta X from a D is equal to v x times t A to 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 gonna be in the X axis, but that's not what we're solving for we're trying to solve for V not and feta. And the way that we would normally solve this is by using vector decomposition. If we want V not X or V X. We would use v not time to the coastline of data. So I'm gonna actually rewrite this Vieques term in terms of V notes and Thetas. So basically what this happened, what with this turns into is I have 45 equals V not times the coastline of theta times 4.5. And if I go ahead and divide this to the other side, I get 10 equals V not times the cosine of feta. Now, unfortunately, there's a problem here is I still have two unknowns in this problem. My Vienna and my theta are both unknown, so I'm stuck here. I can't solve for either of them. So what do I dio whenever I'm stuck? I'm just gonna go to the Y axis on, then try to solve for those variables there. I'm gonna use the same exact interval, the interval from A to D. But now I need my three out of five variables about 9.8 eyes. My sorry negative 9.8 is my a y. Now the initial velocity is gonna be my V a y. Final Velocity is gonna b v d y. And then I've got Delta y from a to D And then I've got tea from a to D. Well, t from a d d 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 m hill, so it's gonna be negative because I'm going downwards. And that's my delta y. So this is negative five. And then what about my V d Y? I don't know what the final velocity is in the Y axis. What about V A. Y? Well, the way that would normally solve this is just like the not xxvi cosine theta my V not y, which is V A. Y is just equal to v knots times the sign of data. So I'm just gonna plug in this expression here for va y. This is just equal to V, not time to the sign of Fada. Now, if I wanted to set up an equation for this, this would be my ignored variable. And even though I don't know Vienna or sine theta, if I wanted to find it, I have my 3 to 5 variables, so I'm just gonna use equation number three which says that Delta Y from A to D is equal to v a Y times t a d plus one half times a y t a d squared. So if I go ahead and plug in everything I know about this problem and then I also replaced my V a y with the not signed data, and what I'm gonna get is I'm gonna get negative five equals V not times with sign of data times 4.5 plus one half negative. 9.8 times 4.5 squared. Now if you bring everything over to the left side. So if you move this term over and the 4.5 over, what you're gonna get is you're gonna get 21 is equal to V, not times the sign of feta. So even though you have everything else in the problem, you still have a problem because you still don't know what Veena r and Theta are. So if you take a look here sometimes what will happen is that for equations or for problems in which two out of these three variables are unknown, you're gonna end up a situation when you're setting up your X and Y equations where you have two equations with the same two unknown variables. Both my X and my Y axis equations. I've ended up with a situation where V not and theta are both unknown. So how do we actually solve for this? What? We're gonna use a math trick, a cool little trick called equation substitution. And the basic idea here is that we're going to solve for one unknown in one of the equations just as an expression. And then we're gonna plug that expression into the second equation. Let me show you how this works. So we've got a situation here and here where we got two equations and two unknowns. So we're gonna use substitution. And really, what happens is if I bring this equation down here, What I wanna do is I want to solve for one of the variables. Now, the easiest one to solve for is gonna be V not because this data is kind of wrapped up inside of the coastline. Term on, it's gonna be hard to get it by itself. So what I wanna do is I wanna move this coastline data term to the left. What I'm going to get is that 10 divided by cosine. Theta is equal to V knots. That's just an expression. Whatever. 10. Whatever coastline data is, if I do 10 divided by coastline data, that's gonna be V knots. So what I do with this expression here is now that I've solved for this unknown variable, I'm gonna plug it into the second equation every time I see a V Knots. So what happens is I'm gonna get 21 equals and then I'm gonna do 10 divided by cosine theta times the sign of data. So notice how Now, in this equation here, I don't have any more V knots. And the only other variable that I have, the only variable that exists is gonna be fate A so I can solve for it. So, really, what happens is this becomes 21 equals 10 and then I have a signed data over a cosine theta. So this is gonna become 10 times tangent data, and this is gonna be 2.1 equals tangent data. So what I can do is say that data equals the tangent in verse off 2.1. So there's gonna be a tangent adverse of 2.1. And what I get is I get my angle, which is 64.5 degrees. So this is now one of my variables. I've solved tour. I've sold for theta, which is 64 4.5 degrees. And so now that's the first variable. So now what I dio is I go back to one of my original my original two equations, which are basically these two equations here, and I can just plug in now that I know tha tha that's 64.5 degrees and I confined. V not you can use any one of those two equations here, so I'm just gonna go ahead and use this one. So 21 is equal to Well, I'm just gonna now divide the sign term to the other side. So this is gonna be 21 divided by the sine of 64.5, which is the number that I'm getting from here. And this is gonna equal v knots. And if you do this, you're gonna get 23.2 m per second. So that is your V not that's your other variable over here. And so what I want you to do now is I want you to take this variable. That's the 64.5 and I want you to solve for me not using that equation, and you're gonna get the exact same number. All right, so we're still just gonna use the exact same steps to solve these kinds of problems. But sometimes you may run into situation, which you have two equations and two unknowns, and you just use equation substitution. That's if this one, guys, let's let's go ahead and get some more practice.

2

Problem

A ball is kicked at a 45° angle from the ground. It hits the wall of a building 30 m away, 10m up from the ground. What was the ball’s initial velocity?

A

17.7 m/s

B

3.2 m/s

C

21 m/s

D

441 m/s