 In my previous examples on projectile motion, I have told you that the range and the trajectory equation is not that important that you can solve most problems by just looking at what's happening in x direction as a function of time and in y direction as a function of time and then you can use time to go from one to the other. In case you ever need the range equation, let's have a look how we develop it and also let's have a look how we develop the trajectory equation. First let's look at the trajectory equation. What is the trajectory equation? The trajectory equation is simply the position in y as a function of x. In order to get it, it's quite simple. We solve the x equation for time and then we plug it into the y equation and then we have eliminated time as a variable. In the example here, we're going to be assuming that x initial is zero and y initial is zero assuming that we're starting from ground going back to ground. So let's see how this looks like. So I've found my time and I'm going to plug it in here and here we want cancels v1. So I can rewrite this as y as a function of x is sine over cosine is 10 so 10 alpha times x minus gx squared over 2v initial squared times cosine squared alpha and this is my trajectory equation. Now the next step is figuring out the range equation. The range being if I throw it from ground to ground, how far did the project fly? In order to do the range equation, I can use the trajectory equation I just got. All I have to do is after I have to plug in the refinal as zero. So zero is equal to 10 alpha times my x minus gx squared over 2v initial squared times cosine squared alpha. Now you see that I have the term x squared. If I'm starting from ground and going back to ground I can actually divide by x. This is something that every mathematician will tell you should never ever do divide by x because x could be zero. In this case let's do it anyway because we know that x equals zero is our initial position. So we know that we're not losing that equation x1 equals zero was one of the solutions we're interested in the other one. So if I divide by x this gets already quite simpler and now all I have to do is solve this for x. So this already is my range equation however it looks a bit ugly so let me simplify this using trigonometric equivalences. Now first of all 10 can be rewritten as sine over cosine. So sine over cosine times 2 times v initial squared times cosine square alpha. Why did I just do that? Because now one cosine cancels out and if I rewrite this I get v initial squared times 2 and sine alpha times cosine alpha over g which if you paid attention is nothing else than sine of 2 theta. So I can rewrite my range equation in the end as v initial squared times sine of 2 theta over g. Note that in both cases you're not going to be plugging in minus 9.8 for g but only 9.8 where we took care of the sine in the step up here. So only put in the value of 9.8 or 10 if you're allowed to round a bit into the trajectory and the range equation. Also please note that these two equations only apply if you start from ground and you're flying back to ground. If you have any other motion you start at some initial height you have to reproduce them based on the equations as a function of time.