Symmetrical Launch

Hey, guys. In this video, we're gonna start looking at projectiles that are thrown upwards rather than flat or downwards, and they were gonna start solving a specific case of this upward launch. But before we do that, I just want to give a general point that whenever you launch something upwards, the initial velocity is always going to be positive. And that's because, for example, in this specific example, here, we've got a football that's launched with some initial velocity at some angle here. So if you break this up into its why its X and Y components, then RV not why is always gonna be positive as well as RV, not X. Now, I'm gonna skip this second point just for now. We're gonna get back to it later on in the video. Now we'll start talking about the specific kind of example that will be solving in this video. So if you have this football that's gonna be kicked up with some angle like this, it's gonna go up to his height, and then it's gonna later returned back to the ground again. So whenever this happens, whenever you have an object that returns to the initial height from which it was thrown or launched, meaning that the UAE final is equal to your white initial. Its trajectory is gonna be perfectly symmetrical. So notice how we can just basically draw a line that splits this parabola in half. So we're gonna start talking about how to solve symmetrical launch problems. And symmetrical launches are a special case of upward launches. Um, and they have some special properties that are gonna make our equations simpler, So let's go ahead and check it out here. All right, So we're gonna get right to the example we've got, ah, football. That's kicked upwards. And we're gonna calculate in this first part time that it takes to reach its maximum height. So before we do that, let's just go ahead and stick to the steps. We're gonna draw the past and x and y and then figure out the points of interest. So if you were only traveling in the X axis, you would basically just go straight along the ground like this and the y axis you would be going up like this and then you were returning back down to where you started from. So what of our points of interest. Well, the first one is just gonna be the initial. That's point A. And then what happens is and it's gonna hit the ground at some later time. But there's something that happens in between, which is it goes up and it reaches its maximum height over here, which is always gonna be a point of interest. This is point B over here like this. So it's gonna go up to point B and then back down again towards point C. So those are our paths in the X and Y axis and our points of interest. So we've got initial final and the maximum height. So now we're gonna figure out the target variable. So let's go ahead and do that. So, in part, we're looking for what we're looking for the time. So that's variable t that it takes to reach its maximum height. So now we just need to figure out the interval and then start working through our equations. So what interval we're looking at? We're looking at a time that it takes to go from the point where it was launched up to its maximum height, which is point B. So this is the interval that we're gonna use for our equations. The one from A to B. Okay, so that means we're looking for t A B. And remember, we're looking for time. The equation we're gonna use first is always gonna be the X axis equation because it's the easiest. So here in the X, we've got Delta X from A to B equals V x times T from A to B. So if we're looking for time that we're gonna need both of these other variables here. So what about the initial velocity or the X axis velocity? Well, that's actually the simplest one, because remember, we have the magnitude. We know that V not is 20 and we have the angle, which is 53 degrees. So RV, not X, which is just V a X, which is just Vieques throughout the whole emotion is gonna be 20 times the cosine of 53. And so you'll get 12. You do the same thing for the y Axis V not why, which is VA Y is just equal to 20 times the sign of 53. And that's 16. So we know what the y velocity or so the X velocity is so we have this. Unfortunately, we don't have the horizontal displacement from A to B. That would be the horizontal distance that's covered from A to B. We don't know anything about that. So, unfortunately, we're a little bit stuck here in the X axis, and so therefore, I'm gonna have to go into the y axis. So in the Y axis, remember, I need my five variables looking for a have a Y, which is always negative 9.8, regardless of the interval that you're using. Um, then I've got the initial velocity, which is just V a y, which I know. 16 the final velocity. It's gonna be velocity at B. Then I've got my delta y from A to B and my tea from a to B. Remember, I came over to this axis here because I'm looking for T A B. So really, in order to solve this equation or just pick one of my equations, I'm gonna need either the final velocity or I'm gonna need the vertical displacement. So, unfortunately for Delta Y from A to B, I don't know what the vertical displacement is from A to B I don't know what the height of that peak is, so I don't know what Delta y from a to B is. But what about V B Y? What's the final velocity? Well, once you're going from a to B here and just like we just what we did for a vertical motion, we can say is that when the object reaches its maximum height on the velocity in the Y, axis is momentarily zero. So think about this projectile as it's moving through the air. It's being, you know, it's why velocity is being reduced by gravity, right? Gravity is pulling this object down. So what happens at point B is its velocity is just perfectly in the X axis. VB X is just equal to V X, and we know that's 12. But the why velocity is equal to zero. In fact, this is a general point that you absolutely should know. The reason that the maximum height is always gonna be a point of interest is because the velocity at that peak in the Y axis he's always going to be zero. All right, so you should know that V b Y r v Peak Y is equal to zero. So means this is zero over here and now we have 3 to 5 variables. Okay, so we're gonna pick the one that that ignores my delta y. And so that is gonna be equation number one. So five final velocity, which is V B Y, is equal to v a y the initial velocity plus a Y times t from A to B. And so we have that this is equal to zero. You know, this is equal to 16 plus negative 9.8 times t from A to B. And so therefore, the time is 1.63 seconds, and that's the answer to party. All right, so that's the answer to part A. Now we're gonna move on to part B. What is partly asking us? Well, party is asking us for the total time of flight, so we go through the steps. We've already have our paths and X and y. Now we just need to determine the target variable. Remember, we're still looking for a time Here is there's still gonna be variable t and now we're just gonna look for the interval. What interval were using? Well before we were just looking for the time from the ground to the maximum height that was ta be. Now we're looking for the whole entire time of flight, so that's actually gonna be from a all the way to see. So that's the interval we're gonna use. That's ta to see here. So we've already got this time here. This ta B is equal to 1.63 seconds. But what I want is that once the entire time of flight, that's t a C like this. Okay, so how do we get that? So if I want Ta C, then what happens is if you look at this diagram the total time of flight here is just gonna be the time that I have already calculated in part a t a b plus the time that it takes to go from the peak back down to the ground again. That's TBC. So I'm basically just set up a simple equation. T a B plus t a c. I already know what this number is. Ta be. It's just 1.63. And so what's the other number? I'm sorry. What's that? That's This should be B two c sorry about that. It should be to see What about the other number? Well, one of the reasons that symmetry is so useful in these kinds of problems is because because we can split this parable a perfectly in half the time it takes to go from the ground up to its peak height. This ta be is gonna be the exact same amount of time that it takes to go from the peak back down again to the original heights. If it went to any other heights when went shorter or higher or something like that, it would not be the same height. But because these things are perfectly symmetrical, then what we can say is the T. V. C is also 1.63 seconds. So therefore the time that it takes for the whole entire time flight is just double what it took to go up. So it's 1 63 plus 1 63 and so therefore, the total time of flight is 3.26 seconds. So for symmetrical launches on Lee, what we can say is that the time up is gonna be the exact same as the time down. That's one of the important questions you should know. But this Onley again works for symmetrical launches. All right, so let's keep going. Sorry about that. So that's the answer. And now let's go ahead and do part c Just gonna have a little bit of just don't have to keep scrolling up and down. So now we're looking for the vertical components off the footballs velocity when it returns back down to the ground again. So when it returns to the ground is gonna be here at point C. So we've got components of the velocity we have. We have a sideways components that's V C X, which is equal to V X, which we know is 12. We've got a Y component V C Y, and we also have the two dimensional velocity V C. If these are the legs of the triangle V, C, X and V C Y, they combined to form a two dimensional vector V c So what are we actually looking for here? We're just looking for the vertical components. So that means we're just looking for V. C Y. So we're looking for VC Y over here, and so which interval are we gonna use. Well, it actually doesn't really matter. We could use the interval from B to C or from a to see. You know, we work the same exact way. Starting to look. We're gonna look at the interval from A to C. And so now, which equation we're gonna use. Well, let's see here. Let's go through a list of variables again in the y axis, we've got a y equals negative. 9.8. We got the initial velocity, Just va y, which is equal to 16. Yeah, so our V c y, or Our final velocity, which is just V C Y. We're looking for Delta y from A to C and then t from A to C. So this is what we're looking for. Our final velocity in the Y axis. So what about Delta y from a. D. C. What is the vertical displacement from the initial to the final? So if you take a look here at this diagram, if we're going up to point B and that back down again to point C again, then that means that our vertical displacement from start to finish is actually zero because we're ending up back at the original height that we started from. So that means the vertical displacement along this specific interval from a D. C. Is equal to zero. And so we also know what the time is. The time from ABC is just 3 26 seconds. So it actually we have four of the five variables we can use any one of our equations to figure out what v C Y is. Whenever this happens, we're just gonna use equation number one because it's the simplest. So one says equation number one says the final velocity V. C. Y is the initial velocity plus a Y times T from A to C. So if you go ahead and do that, R V C Y is gonna be equal to V A. Y, which is 16 plus, then we got negative 9.8 times 3.26 seconds. And if you work this out, what you're gonna get, you're gonna get negative 16. So what does that mean? That means that the y component of your velocity here the ground is gonna be negative. 16. Now, notice how this is basically just the same exact number is just the opposite sign of what you started off with here. You started off with 16 upwards and then you ended up with 16 going downwards. This is no coincidence. This is actually in fact, another consequence of symmetry. So for symmetrical launches on Lee, the time up is gonna be equal to the time down. But the velocity up is also going to be equal to the negative. Off the velocity down, they have the same number, just opposite signs. So v c Y here is gonna be equal to negative V a y. Now, one of the other consequences of this is we know that this initial velocity is 20. So we know that basically V not or via is 20 because of the legs. We have 16 and with this is 12. Now, when we get to part C over here, we have the exact same numbers except one of them is negative. But when we re combine this into the two dimensional vector here, V C is also just going to be 20. So we can say here is that the magnitude of V C is equal to the magnitude of the A. The initial velocity is gonna be the same as the final velocity. And because their magnitudes, it's just the numbers, they're gonna be the same. However, the angle stay to see here is just gonna be the opposite of data. A. So now what happens is this angle goes below the X axis. And so this is gonna be negative. 53 degrees. Alright, guys, that's it for this one. Let me know if you have any questions.

Hey, guys, let's check out this problem. We're playing catch on a far away planet. We've got some information about the launch speed and angle of a projectile and the distance that it travels, and we're gonna calculate the gravitational acceleration on this planet. So let's just stick to the steps. Basically, what I've got here is we've got a level ground. I'm told that I'm throwing a ball upwards with some velocity at degrees, and then it's gonna later return to the ground again. So that means when I draw out this trajectory, it's actually just going to be a symmetrical launch because it's gonna launch. It's basically gonna land the same things like height that it started from. So so this is basically a diagram. The first step is we want to break this up into its X and Y paths, draw them and then draw the points of interest. So in the X axis, we would just go straight from here to here, passing the maximum height, which is always gonna be a point of interest right there. So and in the Y axis, we're just gonna go up and then come straight back down again. So this is gonna be our initial points. Then this point here where a maximum height is, is gonna be be. And then finally, we're gonna hit the ground again at point C. So these air points of interest here on the Y axis will be going a from a up to be and then back down to see again. So that's this first step Second step is we're gonna determine the target variable. What are we looking for in this problem? We're looking for the gravitational acceleration. So remember that that variable is little G. So where do we see little G and our equations? Well, in our Y axis equations, little G is always inside of this A Y term. So we're really looking for when you were asking for the gravitational acceleration because it says magnitude. All these numbers gonna be positive here, But we're actually really looking for a y here. So that's gonna be our target variable so that the next step is what interval we're gonna look at. Well, here's where you actually have some options because in symmetrical launches, remember, we have a bunch of special properties about symmetry we can use a to be or you can use a to B to C s so you can always just choose one interval. And if it doesn't work, then we'll choose a different one. So let's go ahead and choose the interval from A to B. So if we choose Interval from A to B and we're looking for a y axis variable that we're gonna have to go ahead and list all of our variables out groups. So in the y axis, I've got a Y, which before we just said was equal to negative G. However, because we're not on the earth, we can't say it's negative. 98 That's the whole point of this problem is the gravitational acceleration off this different planet here is not going to be 9.8. So don't make that mistake s always be on the lookout for these kinds of problems. We could always assume 9.8 if we're on the earth. Okay, so this is actually where we're gonna be looking for. So what are other variables? RV Not y is va y V Our final Why is gonna be V B y? Anyway? You've got Delta y from a to B, and then you've got to tea from a to B. Okay, so what about our initial velocity in the Y axis? Well, let's see, we've got a launch velocity. This is V Knots or V A. We know this is equal to 10. We also have the angle is 37 degrees. So we can actually break this up into its X and Y components V A X, and then via Sorry via X and V A y. And let's see, R V A X is just gonna be, um, the VX throughout the whole entire motion. That's gonna be 10 times the co sign of 37 which is eight. And then our v A y is just going to be 10 times the sign off 37. And that's gonna be six. So you know what? Our initial velocity is in the y axis. That's just six. What about the final velocity in this interval? So, basically, we're only just looking at this interval right here from a to B. So we start off with, uh, 6 m per second in the Y axis. And then what do we end with when we get to point beat will remember that in vertical motion or projectile motion. The reason that this maximum height is very useful is because VB Y is equal to zero. We know that the velocity, once it gets up to its peak, is gonna be momentarily zero there. So we know that this is gonna be zero. And so what about Delta y from A to B? Well, that's gonna be the distance, the vertical distance from A to B. And we don't know that, Um, what about time? So do we know time? We also don't know the time either. So it looks like we're kind of stuck here. We don't know the time and we don't know Delta y from A to B off these two variables, the one that I could solve for by going to the other equation is gonna be the time. Remember where we were stuck in the Y axis would go to the X axis, and I can't solve for Delta y from A to B in the X axis. So instead, what I can dio is I consult for time, So let's go ahead and do that. Then when you go to the X axis and I want t a to be so I'm gonna use the equation. Delta X from A to B is equal to v x times t from a to B. Um, now what happens is what is the Delta X from A to B well, the Onley Delta X The only horizontal displacement that I know is I know that the horizontal displacement from A to C is equal to 32 m. So I know this whole entire thing here. So let me see if I can write this. This whole entire thing is 32 m, but from symmetry when when things we can do here is we could basically cut this projectile motion basically in half. Then that means if the whole entire horizontal displacement is 32 then each one of these pieces here is 16 m from a to B. It's 16 m and from B to C at 16 m. So it means that Delta X from A to B is 16. My ex velocity here is eight and ta be eyes my unknown variable. So that means if I just go ahead and saw for this, my t A B is equal to two seconds. So now I can actually plug this back in here. I know. That's two seconds now. And so now I have three out of five variables, and I could go ahead and pick the equation that ignores Delta. Why? So if I go ahead and do that, that's just gonna be equation number one. And it says that the final velocity V b Y is equal to my initial velocity v a y plus a Y times t from A to B. I know this is gonna be zero freebie y zero. I know this is gonna be six plus and then I've got a Y times two over here. So if you go ahead and solve for this, what you're gonna get is that a Y, which is equal to negative G is just negative. 3 m per second squared. So what does that mean? It means that the magnitude of the gravitational acceleration is 3 m per second squared. And so if you go to our answer choices, that is answer choice. See? Alright, guys, that's it for this one