 When learning projectile motion, a common example to discuss is to look at the range of a projectile and what things may affect that. And if you are kicking or hitting a projectile at what angle of launch will it travel the furthest? So let me draw a picture of our scenario. Our projectile is launched with an angle of v initial. Sorry, with an initial speed, launching speed of v initial and a launching angle of theta. So our goal is to find an expression for the range as a function of these two initial parameters. And to use that information to figure out what launching angle will give us the maximum range. So I'm just going to add a label for our horizontal displacement here. We will call that the range. And I'm going to write out our goal. Our secondary goal for this will be to find what angle maximizes the range. So let's take a look at what we know given our setup here. We are looking at the special case where the projectile is launched and lands at the same vertical position. So it will start at zero and it will finish at a height of zero. As for any projectile problem, it's very important for us to identify our coordinate system since our velocity, displacement, and acceleration are vector quantities. So I will draw a coordinate system defining up to end to the right to be positive. So now as with any projectile problem, what we need to do is we need to look at the horizontal and vertical components of motion independently. There it is the vertical acceleration that will determine the how long it takes the ball to rise and fall back down to its original height. And it is the horizontal component of motion that will determine how far it travels during that time. But these two components of directions of motion can be analyzed independently. So let's go ahead and write down what we know. So I'm going to set up two columns here in which we can identify our horizontal information and our vertical information. Let's start with our horizontal direction. We know that projectiles do not accelerate horizontally. We know that our displacement is going to be our change in position. So we are starting here horizontally and we are finishing here horizontally. We know that our v-initial is going to be related to our launching speed and angle. And we know that our time for both the horizontal and vertical components of motion will be the same. So let's write down our displacement and let's figure out how we're going to deal with our velocity. I'm going to sketch for us down here our initial velocity vector and its horizontal and vertical components. This purple vector represents our initial launching speed. This here is our angle theta. So this will be our v-initial in the vertical direction. And this blue vector is representing our v-initial in the horizontal direction. As you can see from the right triangle that both the vertical component will be related to the angle theta and to the v-initial by the sine function, whereas the horizontal component will be related to those by the cosine, related to those by the cosine function. Let's go ahead and finish writing down what we know in our horizontal direction. And in our vertical direction we know that our acceleration is going to be, have the magnitude of the acceleration due to gravity and because I chose up to be positive and the acceleration due to gravity is a vector in the downward direction, this will be negative. We have displaced zero in our vertical direction because we leave and return to the same vertical position and displacement is the change in our position, in this case in our vertical position. We have already written down an expression for our v-initial vertical and we know that our vertical motion will last for the same amount of time as our horizontal motion. So what we want to do is using the things that are common here such as our v-initial and our time to figure out an expression for our range in terms of our launching speed and our angle theta. So our goal in the algebra will be to find a common variable that, such as time, that is common for both the vertical and horizontal components of motion and write down expressions for those and then we can set those equal to each other. So the next few steps will show how we find, use our horizontal information to find an expression for the time and then we're going to substitute that into an expression for our vertical motion and we will see where that takes us. Let's write down the equation of kinematics for our horizontal motion. If we look at our equations kinematics we'll see that one of them contains the acceleration, the displacement, the v-initial and the time and if we substitute what we know this will simplify. The last term here does not have to be included because the acceleration is zero. So here we have range in terms of our v-initial and our cosine theta but we have time. We don't want our final expression for range to have time in it so I may have misspoken before when I said that we would solve for time in the horizontal direction then plug it into a vertical expression. We're actually going to do the opposite of that. What we can do is see if we can use our vertical information to find an expression for the time that will allow us to eliminate the variable time from our expression. I'm just going to clear up a little space here and go ahead and start that calculation. As you can see with the variables that we know we would want to use the same equation that we did in the horizontal direction. If we plug in what we know and we can simplify this a little bit we want to solve this equation for t. By doing our algebra here we can get an expression for t of 2 times v-initial sine theta over g. Let's take this expression for t and plug that into our range equation. So we're going to take this and we're going to plug it in right there. And we're going to simplify this a little bit further because as it looks right now it doesn't look to be a very simple expression for the range as a function of our angle theta. But what's going to help us is a trig identity. Trig identities are not something that most students are expected to learn at some point but may not commit to memory for the long term. So I'm going to share with you one of the double angle formulas that 2 times sine of theta cosine of theta is equal to sine of 2 theta. In most math notebooks you can look this one up. Again it's not something that most people have committed to memory but of course you are welcome to reference as necessary in order to try to make things simpler and find a better solution to our problem. So I'm going to substitute this in to our range formula. So now we can see we have derived an expression for our range. That depends fairly simply on our launching speed v-initial, our acceleration due to gravity. Most times we'll assume we're doing this on the earth and the angle theta. From this you can see that so we've met our first goal which is to find an expression for this. Our sort of secondary goal was to see if we can use this information to figure out where what angle would give us the maximum range. So if we have a fixed launching speed v-initial here and g are going to be constant. So our range is going to depend on the sine of 2 theta. From your trigonometry studies you should know that the maximum that a sine function can be is 1. So we know that the sine of 90 degrees is equal to 1 and 1 is a max value for a sine function. So 90 degrees gives our sine, gives us a maximum for this sine function. So at 90, if the maximum for this here is 90 degrees, alright so our max range is when 2 theta will be equal to 90 degrees therefore when theta is equal to 45 degrees. This answer should make you feel pretty good because it seems rather intuitive. This is an intermediate angle between firing something straight up and firing it perfectly horizontally. We can, so if you imagine at this angle we will be able to get enough vertical motion to keep the projectile in the air for a sufficient amount of time and enough horizontal velocity to allow it to travel for a fair distance. This example was the simplified version in which we launch and return to the same height. There is a more complex version in which the launching height, the launching vertical position and landing position are different and that's a slightly more complicated derivation but it will use the same idea and that's something for a future lesson. So hopefully this has helped you learn a little bit more about the range of a projectile.