Hey, guys. So in the last couple of videos, we saw how to solve problems in which an object is launched upwards and then returns to the same height from which it was launched. This was called symmetrical launches, but this won't always happen. You might see some problems where objects are launched and they return to different heights. There are really just two options. You could land at a higher height or a lower height. 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. We're really just going to use the same steps and equations as before. There's just a couple of new things here. So what would this look like if you landed at a higher height, 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 wall or a building or something like that. And if you were launching at a lower height, I like to call these launch from height problems, then it would look like this. It would go up, reach its peak and then come back down again and then hit the ground. Alright?

So before we get into this next bullet point, I actually just want to go ahead and start the problem because we're just going to use the same exact system to solve them. So we've got this potato. We're going to fire it from a potato launcher while we're on top of a cliff, which is 20 meters high. And so before we go ahead and draw anything and start working out the variables and equations, we're just going to solve the first step. We're going to draw the paths in x and y. We've got this object here, and it's going to go up and down like this. If we could only move along the X-axis, it would look like this, and the Y-axis would go up to the peak and then come back down again towards the bottom. So what are our points of interest? Remember, we're 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 which happens in between. So what happens is this object is going to go come back around again. It's going to pass the heights from which it was initially launched and then continue further. So this is going to be b. We'll call this 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 going to be just like a symmetrical launch problem. 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?

Again, we're going to get back to the second bullet point right here. Let's just get right to the problem. So we've got the paths in x and y. Now we're going to determine the target variable. What we are actually looking for here is the vertical component of the potato's velocity just before it hits the ground. So that's not going to be point D. Remember, there are two components of the velocity. There's going to be the x component, which I'll call v_{dx}, which is just v_{x} because it never changes, and then we're looking for v_{dy}. So this is what we are looking for here. Remember these two components combine to produce a two-dimensional vector v_{d}, but that's not what we're looking for. We're just looking for v_{dy}. So we're looking for that target variable here.

Now, the next big question is: what interval are we going to use? Remember, we have a bunch of different points of interest now. There are many different options as to the interval that we could choose. There are a couple of options here, and depending on which one you take, you can still get the right answer. But there are a couple of options that are easier and harder. One thing you might be thinking of is if you have the initial velocity, which is v_{a}, which is just equal to 30, then what you could use is a symmetry argument. You could say, well, if v_{a} is 30, then I know that v_{c} is also going to be 30, except it's just going to be pointing in the opposite direction. So I could use my components v_{cx}, which is v_{x}, and then whatever v_{cy} is is just going to be the negative of v_{ay}. So you could use this interval just from c to d, and you absolutely could solve it that way. But I'm going to warn you that there's going to be a bit more work, and you're going to have to solve a bunch of other things. The easiest choice for these kinds of problems and, in general, what you should try to do if you're ever landing at a lower height is b, which is the maximum height. You're always going to want to try to include this in your equations because one thing that we know is that the y velocity when it's at the peak is equal to 0, and we're going to see how this simplifies our equations and our variables. Alright? So we're going to be looking at the interval from b to d. So we're looking for the y-axis, so I need all my y variables. So I've got my acceleration in y, which is always negative 9.8, the initial velocity. Well, the initial velocity in this interval from b to d is actually going to be v_{by} and that's 0. This is why this interval is really easy to use and you should always try to use it because we've already unlocked. We've already got two out of the five variables that we need. We just need one more, and then we'll be able to solve our equations. So the final velocity is going to be at point d, which is exactly what we're looking for. And then we've got Δy 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 Δy 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 the potato is launched from a 20-meter-high cliff. So is that the Δy from b to d? Well, no, because that's only going to be this piece right here. This vertical distance is the 20, not this whole entire thing over here. However, what we're also told from the problem is that the potato reaches its maximum height of 49.4 meters above the ground. So this is going to be our Δy from b to d. However, because this points downward, it's going to pick up a negative sign. So that's Δy, which is going to be a 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've already got 3 out of 5 variables here. Alright? So we're just going to use the one that ignores time, that's equation number 2. So the final velocity which is v_{dy} squared is going to be the initial velocity v_{by} squared plus 2a_{y}Δy from b to d. Okay?

So we already know that the y velocity v_{by} is going to be 0. And so v_{dy} squared is just going to be 2 times negative 9.8 times negative 49.4. And so if you go ahead and work this out, we're going to get that v_{dy} is equal to the square root of 960, which is just going to be approximately 31.1. However, there are two answers here. We could use 31.1, which is going to be a positive 31.1, which means that we're getting a velocity that points up, or we could get a negative 31.1 because remember we're taking the square root of a number here and we're solving for velocity. That would correspond to a velocity that points downwards. So which one of these answers is correct? Well, if we take a look at v_{dy}, it points downwards. So that means that the positive answer is not the correct one. Instead, we're going to use v_{dy} is equal to negative 31.1 meters per second. 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.