Hey, guys. So in the last video, we were introduced to symmetrical launches, which is a type of projectile motion in which the final height equals the initial height. We saw some special properties of those kinds of motions. The fact that t_{up} = t_{down} and we saw that v_{up} = -v_{down}, various things that come from symmetry. But on top of these special properties that apply to symmetrical launches, there are also some special equations that you may be allowed to use if your professor allows you to, and these are equations that are going to make symmetrical launches even easier to solve. So I want to show you how that works in this video. We're just going to get right to the example.

We've got a catapult that launches a projectile. We've got the launch speed and the angle. We're told that this v_{0} here, which is just VA, is equal to 100, and we're told that the launch angle θ_{a} is 53 degrees. And we're going to find in this first part here the time that it takes for the projectile to hit the ground again. So what that means is we have to figure out the time that it takes for the projectile to go from point A to point C. Now the way that we did this before in the last video was that we basically just broke this up into 2 symmetrical pieces. We found what the time up was and then we saw that the time up equals the time down and so we were able to just double it. However, a more straightforward approach is actually just called the total time of flight equation. This is one of the special equations that we'll be covering, and it's basically just 2v_{0}sinθg. All you need to know is just the initial velocity and the launch angle, and you can actually figure out the time of flight for a symmetrical launch. This only works for symmetrical launch equations and problems, so just make sure you understand that.

This equation actually is a consequence of one of our symmetry, equations like v_{up} = -v_{down}. I want to show you this really quickly. In the A to C interval here if you use equation number 1 from your UAM equations, you have that the final velocity v_{cy} is v_{ay} plus a_{y} times t_{ac}. But what we know from symmetry is that if you have the initial launch velocity, if you have the y component, then when it gets to point C, it's going to be the same exact number except it's going to be downwards and so it's going to be negative. So v_{ay}, or sorry, v_{cy} is equal to negative v_{ay}. So if we go back to this problem here, we can actually just replace this v_{cy} with negative v_{ay}, right, because of this property over here. And this is v_{ay} plus a_{y} t_{ac}. So this actually just going to become 2v_{ay}-g. So that means that t_{ac} is just Well, what is v_{ay}? Well, if you have the launch velocity and you have the angle, then v_{ay}, v_{ax} is just v_{0} cos θ and v_{ay} is just v_{0} sin θ. So that means that our equation just becomes 2v_{0}sinθg, and this is actually just the equation that we just, we just arrived at over here. So what we can use is we can just use this equation straight from the get-go and say the t_{ac} is 2×100×sin(53°)9.8. That gives us the entire time of flight. We don't have to break this thing up. And so if you actually plug this in, you're going to get 16.3 seconds, and that's the answer. So that is the answer to the total time of flight. So let's move on now to the horizontal range of the projectile.

So what does that mean? Well, the horizontal range is, another word for that is actually just the total horizontal displacement. It's how much distance the projectile covers from start to finish. So just like this whole entire interval here from A to C is the whole time, then the horizontal range is just going to be R which is equal to Δx from A to C. It's going to be the entire distance that the horizontal that the projectile covers horizontally. So R which equals Δx from A to C. If we were to set up our equations, the only equation that we use in the x-axis is just v_{x} times t from A to C. So what we can use is now that we've actually figured out this time, the 16.3 seconds, we can actually just go ahead and plug that straight into the equation. So in general, what happens is if you do have the total time of flight, if you have t_{ac}, then your range equation is just going to be Δx from A to C. It's just going to be velocity times time. It's basically just the normal, you know, equation that we've been seeing so far. However, if you didn't already calculate that total time of flights, then there's actually another equation we can use. It's called the range equation. And so R, which is still equal to Δx from A to C, just has a different form and it's going to be v_{0}2×sin2θg. So what I want to do in this problem is I want to show you that using both of these equations will still get the same exact number.

So for instance, what is the x component of the velocity? Well, we'll go back here and say that the x component of the velocity is going to be v_{0} times cos θ. So it's going to be 100 times the cosine of 53. And if you go ahead and do that, what you should get is you should get 60. So if we plug it into our problem here, we've got 60 times and then we've got the t_{ac}, that's 16.3. So that's just going to be 980 meters. So that's what we get for the range. However, if we didn't already pick out the time of flight, and if we only just had the launch angle, the launch velocity, and the angle, then we could just plug it straight into the range equation. And so what we would do is we'd have to plug in the magnitude of the initial velocity, not the components. So this would be 100 squared times the sine of 2 θ. That's 2 times the launch angle. So we're going to do 2 times 53 degrees and then divided by 9.8. That's g. And if you plug this in, you're actually going to get 980 again. So we're going to get the same exact answer, and that's because really both of these equations mean the same thing. They're equal to each other. Alright? So that's how you use this range equation.

Here's just, you know, so you can memorize it or you use it on your homework and test questions and stuff like that. So there are a couple of things you should know about this range equation. The first is that it's maximum when the launch angle equals 45 degrees. So remember that this equation only depends on the initial. So what that means here is I've got this graph of these projectiles that are all thrown with the same exact v_{0}, the same velocity, and the one that goes the farthest is going to be the one where it's launched at 45 degrees. Anything less than or greater than 45 degrees is going to give you a smaller range, which actually brings me to my second point. Complementary angles, meaning θ_{1} and θ_{2} equal to 90 degrees, such as 30 and 60, right, because 30 plus 60 equals 90, complementary angles will give you the same exact range for the same exact initial velocity. So again, if all these projectiles are thrown with the same v_{0}s, then what happens is if I were to throw this projectile at 30 degrees rather than 45, then it would go it wouldn't go as high and it would also fall a little bit shorter than the maximum range at 45 degrees. If I were to instead launch this at 60 degrees, it would go a little bit higher, but it would also fall to the same exact range as the 30 degree angle over here. So that brings us to the last part where you're going to find the other launch angle that gives the same exact range as before, the 980 meters. So if we already have one of our angles and we know that θ_{1} is equal to 53 degrees, and θ_{2} is just going to be 90 minus θ_{1}. So it's going to be 90 minus 53 degrees, and that's going to be 37. The way you can test that is you can just plug R you can plug in θ equals 37 into your range equation, use the same exact numbers, so we're still going to use the same 100 squared times the sine of 2 times 37 now. If we divide this by 9.8, we're going to get 980 meters. So we're going to get the exact same range even though the launch angle is different, and that's because they're complementary angles. That's it for this one guys. Let me know if you have any questions.