Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Solving Non-Symmetrical Upward Launch Problems

Patrick Ford
Was this helpful?
Hey, guys. So in the last couple videos we saw how to solve problems. It was an object is launched upwards and then returns to the same height from which was launched. This was called symmetrical launches. But this won't always happened. You might see some problems which obviously launched and they returned two different heights. There's really just two options You could landed a higher height or a lower height here. And so in general, if an object is launched upwards and it lands at a higher or lower height, then the motion is going to be not symmetrical. So that's what I want to show you in this video. How to solve non symmetrical upward launch problems were really just gonna use the same steps and equations as before. Um, this just a couple of new things here. So what would this look like if you were landed at a higher heights? So this trajectory would be like this. It would go up and then before coming back down to the original height, it would hit its target or a wal or a building or something like that. And if you were launching at a lower height, so I like to call these launch from high problems. Then it would look like this. He would go up, reach its peak, and then it would come back down again and then hit the ground again. All right, So before we get into this and next bullet point, I actually just want to go ahead and start the problem, because, really, we're just gonna use the same exact system to solve them. So we've got this potato we're gonna fire from a potato launcher. We're on top of a cliff. It's 20 m. And so before we go ahead and draw anything on, start working their variables and equations, we're just gonna solve the first step. We're gonna draw the path C, X and y. So we've got this object here and it's gonna go up and down like this. So if we could only move along the x axis, it would look like that. And the Y axis. We'd go up to the peak, they would come back down again towards the bottom. So what are points of interest? Remember looking for things like the initial and the final and the maximum height. So this is our initial and then we've got the maximum height over here and we've got the final, which is where we hit the ground. But there's one other point of interest that happens in between. So what happens is this object is gonna go come back around again. It's gonna pass the heights from which it was initially launched and then continue on further. So this is gonna be be we'll call the C and then D. So what happens is if you're ever launching and landing at a lower height, then what happens is part of the motion from A to C is going to be symmetrical. So we see how from A to C, this is gonna be just like a symmetrical launch problem. But what's unique about these kinds of problems is that the object also continues to drop further in this interval from C to D. Alright, so again, we're gonna get back to the second the second little bullet point right here. Um, let's just get right to the problem. So we've got the paths and X and y. Now we're gonna determine the target variable. What are we actually looking for here? We're looking for the vertical component of the potatoes velocity just before it hits the ground. So that's gonna be at point D remember, there's two components of the velocity. There's gonna be the X components. I'll call this VD X, which is just the X because it never changes. And then we're looking for V. D. Y. So this is gonna be we're looking for here. Remember, these two components combined to produce a two dimensional vector v d. But that's not what we're looking for. We're just looking for V D. Y. So we're looking for that target variable here. Now the next big question is, what interval are we going to use or remember we have a bunch of different points of interest now. There's lots of different options as to the interval that we could choose. So there's a couple of options here, and depending on which one you take, you still get the right answer. But there's a couple of options that are easier and harder. One thing that you might be thinking of is, if you have the initial velocity, which is V A, which is just equal to 30 and what you could uses, you could use a symmetry argument you could say, Well, if VA is 30 then I know that V. C is also going to be 30 except just gonna be pointing in the opposite direction so I could use my components V C X, which is V X and then whatever V. C Y is is just gonna be the negative of V A y. So you could use this interval just from C to D, and you absolutely could solve it that way. But I'm actually gonna warn you that there's gonna be a little bit more work on. You're gonna have to go and solve a bunch of other things, the easiest choice for these kinds of problems and in general, what you should try to dio if you're ever landing at a lower height, is you try to choose intervals in which you're trying to include point B, which is the maximum height. You're always gonna want to try to include this. Equated this into your equations because one thing that we know is that the UAE velocity, when it's at the peak, is equal to zero, and we're going to see how this simplifies our equations and our variables. All right, So we're gonna be looking at the variable at the interval from B to D. So we're looking for the y axis, So I need all my wife variables. So I've got my A y, which is always negative 9.8 the initial velocity. Well, the initial velocity in this interval from B d. D is actually gonna be V b y. And you know that. Zero. This is actually why this interval is actually really easy to use. And you should always try to use it because we've already unlocked. We've already gotten two out of the five variables that we need. We just need one more, and it will be able to solve our equations. So the final velocity is gonna be the velocity at point D, which is exactly what we're looking for. And then we've got Delta y for me to be our sorry Delta y from B to D and then t from B to D. So we just need one of these other variables here. Let's take a look at Delta y from B to D. So what does that mean? That would actually represent the vertical displacement from point B, All the way down to the ground like this. So this vertical displacement over here. So which number is that? We'll see if we take a look at the problem. We're told that the potato is launched from a 20 m high cliff. So is that the Delta Y from ability? Well, no, because that's only gonna be this piece right here. This vertical distance is the 20 not this whole entire thing over here. However, we're also told from the problem is that the potato reaches its maximum height 49.4 m above the ground. So this is the number that this line represents 49.4 m. So this is gonna be our Delta y from B to D. However, because this points downward, it's gonna pick up a negative sign. So that's Delta Delta y. That's gonna be 49 point. Whoops. It's being negative. 49.4, and then we don't know the time. We don't know how long it takes to go from the peak back down to the ground again. But that's okay because we already have three out of five variables here. All right, so we're just gonna use the one that ignores time. That's my ignored variable. And that's gonna be equation number two. So this is the final velocity, which is V D. Y squared is gonna be the initial velocity V b Y squared. Plus, uh, a Y times t uh, I'm sorry. This is gonna be plus to a y Delta y from B to D. Okay, so we already know that this y velocity is gonna be zero B B y. And so v d Y squared is just gonna be two times negative. 9.8 times negative, 49.4. So if you go ahead and work this out, we're gonna get that V d. Y is equal to the square roots off, and this is gonna be 900. Um, this is gonna be 960 which is just gonna be 31.1. However, there's two answers here. We could use 31.1, and it's gonna be a positive 31.1, which means that we're getting a velocity that points up. Or you could get a negative 31.1. Because, remember, we're taking the square root of a number here. We're solving for velocity that would correspond to a velocity that points downwards. So which one of these answers is correct? Well, if you take a look at DT y ved white points downwards. So that means that this answer is not the correct one. And instead, we're gonna use V. D. Y is equal to negative 31.1 m per second. That's the That's the vertical component of the velocity right before it hits the ground. That's it for this one. Guys, let me know if you have any questions.