Okay class, so let's continue with the Apollo problem. So everything that we did before in the previous video, in general you can use it for any problem. Essentially what we did was to give the kinetic equation for the position in X and in Y to calculate to derive the range. So for how long an object will go, it's a function of the angle that it makes with the horizontal, the initial velocity and the acceleration to gravity wherever you are. So if you already made this formula, then the problem is much shorter, right? So just want to know the difference between the range on the moon and on the earth. So the conditions will be the same. This is the acceleration that we have in the moon. You should be losing the same sign of the data is the same. Whenever you are solving a problem, you should try to go as deep as possible. We integral or we algebra, so keep your very most in there until the very very end. It's a more elegant way to solve problems, but you know, so many mice is mistakes because sometimes you have, you know, 7.35 here. And then something over here, you multiply, you get a different number that you don't know where, you know, easily you don't know where it came from. But if we keep all your variables in here until the variant, then it is much, it is more likely that you're going to get the right solution. So over here we have the range on the moon. Here we have the range over earth. To have a oxygen to gravity on the moon and on earth. And these two are the same. And we know the acceleration to gravity on the moon is 1-6th that on earth. So that's kind of a nice equation. This is going to be... Right. So we can move the seats up here. And so these becomes... Right. So we can get rid of these part over here and this G over here. So we can make it look even nicer. We can move the fly over here and just divide everything by the acceleration to the gravity on earth. And so now you can plug in your numbers so you know, be not. enforce the values 27. And if I measure orders, I will measure the speed by the dash at the fourth, and see the gray. And then I will take a colour that looks like 0.2 degree to sharp it and get that negative. And we will get the red wind. And we take the waw換 of the euros you need. So, I'm going to cut you later. I get that the difference in the ranges is 276. I get that the difference in the ranges is 276.15 meters. I get that the difference in the ranges is 276.15 meters. This is meter squared. This is meter and second is squared. You can see it, but it gets live. You can get that with your meter. There you go. That's part eight. I'm going to play over here. The last part of the problem asks for the difference in time. For how much longer is the ball airborne on the moon? I guess there's no air on the moon. Well, whatever. Compared to the earth. We already have the answer. It's over here. So, again, the beanets are the same. The same. That is the same expression that we had before. The last five. Over G.E. Take the five out. It's 10. We're not sine theta over G.E. So, 10. 25 meters per second. Sine theta. Sine of 30 degrees. We're here. 0.5. We're divided by 9.8. 10. I've got 12 points in 5 seconds. That was the Apollo problem. We were able to derive the range. I feel free to use that formula. Sometimes you might be asked for things that are part of the derivation. I'm not sure if you're referring to that. But, it's not part of the range.