Hey, guys. In this video, we're going to start looking at projectiles that are thrown upwards rather than flat or downwards, and then we're going to 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 this is 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 \( y \) and \( x \) components, then our \( v_{0y} \) is always going to be positive as well as our \( v_{0x} \). Now, I'm going to skip this second point just for now. We're going to get back to it later on in the video.

Now let's start talking about this specific kind of example that we'll be solving in this video. So if we have this football, that's going to be kicked up at some angle like this. It's going to go up to its height and then it's going to later return 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 \( y \) final is equal to your \( y \) initial, its trajectory is going to be perfectly symmetrical. So notice how we can just basically draw a line that splits this parabola in half. So we're going to start talking about how to solve symmetrical launch problems. Symmetrical launches are a special case of upward launches, and they have some special properties that are going to make our equations simpler.

Alright. So we're going to get right to the example. We've got a football that's kicked upwards and we're going to calculate in this first part the time that it takes to reach its maximum height. Before we do that, let's just go ahead and stick to the steps. We're going to draw the paths in \( x \) 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 would have returning back down to where you started from. So what are our points of interest? Well, the first one is just going to be the initial, that's point \( a \). And then what happens is, and it's going to 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 going to be a point of interest. This is point \( b \) over here like this. So it's going to 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 going to figure out the target variable. Let's go ahead and do that.

In part \( a \), we're looking for the time, so that's variable \( t \) that it takes to reach its maximum height. Now we just need to figure out the interval and then start working through our equations. The interval we are looking at is the time that it takes to go from the point where it just launched up to its maximum height, which is point \( b \). So this is the interval that we're going to use for our equations, the one from \( a \) to \( b \).

Okay. So that means we're looking for \( T_{AB} \), and remember we're looking for time. The equation that we're going to use first is always going to 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, then we're going to need both of these other variables here. 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 \( v_0 \) is 20 and we have the angle which is 53 degrees. So our \( v_{0x} \) which is just \( v_{ax} \), which is just \( v_x \) throughout the whole motion is going to be \( 20 \times \cos(53^\circ) \), and so you'll get 12. You do the same thing for the \( y \) axis, \( v_{0y} \) which is \( v_{ay} \) is just equal to \( 20 \times \sin(53^\circ) \), 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 going to have to go into the \( y \) axis. So in the \( y \) axis, remember, I need my five variables. I'm looking for \( a \), I have \( a_y \) which is always negative \( 9.8 \) regardless of the interval that you're using. Then I've got the initial velocity which is just \( v_{ay} \), which I know is 16. The final velocity, it's going to be velocity at \( b \). Then I've got my \( \Delta y \) from \( a \) to \( b \), and my \( t \) from \( a \) to \( b \). Remember, I came over to this axis here because I'm looking for \( T_{AB} \). So really, in order to solve this equation or just pick one of my equations, I'm going to need either the final velocity or I'm going to 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 \( V_{by} \)? What's the final velocity? Well, once you're going from \( a \) to \( b \) here, then just like we just what we did for vertical motion, what we can say is that when the object reaches its maximum height, then the velocity in the \( y \) axis is momentarily 0. So think about this projectile as it's moving through the air, it's, being, you know, it's \( y \) velocity is be