 Okay, so another projectile motion example. I posted one already, excuse me, but let me just go ahead and do the most complicated problem that you would probably see. So let me just go ahead and set it up, and I'm just making up stuff as usual. So let's say I shoot a ball off a cliff, which I did before, but now what if I shoot it off the cliff at an angle? So now let me just make up some things, shoot it at an angle of 35 degrees, totally made that up with a velocity of five meters per second. And let's say the cliff is two meters high, or it's a table, it's a tall table, it's a meter high, it's a tall table, it's like a bookshelf. There. Okay, so clearly the question is where is it going to land? Okay, so it's going to do something like this. So what are we starting with besides the initial velocity, the magnitude, the angle, and the height? Let's write this down in a way that we can say things. Let's say Ay equals negative 9.8 meters per second squared, because it's projectile motion. If air resistors doesn't matter, then that's fine. So I also need the x zero, I'm going to call this, I'm going to call this the origin. So x zero equals zero, x final equals, we don't know, we're trying to find that. Y zero equals two meters, and then y final equals zero. Now I also need the initial x and the initial y velocity. So vx zero, if I have a velocity like that, and I know the angle, this is vx, that's vy, so vx is going to be the total magnitude times the cosine of that angle, this is going to be five meters per second times cosine of 35, and then the y velocity initial five meters per second times the sine of 35. Okay, so when we're dealing with projectile motion, since the velocity and the acceleration, the x direction, which is actually ax, you can write this as zero if you want, is zero and they don't relate. We actually have two one-dimensional problems that just happen to have the same time. So in the x direction with no acceleration, I can say x final equals x initial plus vx. And if I start at time t equals zero, which I will, just to be lazy, I want to say t. I don't have to say vx zero or vx, the x velocity never changes. So I can just call it vx, it's just one value, but that's one thing I know, okay? So I know the initial x, I know the x velocity, but I can't find where it lands because I don't know how long it took, okay? So if I can find out how long, then I can find out how far it went. Okay, what about the y direction? Well here I know y equals y zero plus v y zero t minus one-half gt squared. So can I use this to find the time and then put that back into the x direction to find the displacement or where it lands? I can, okay? Because here I know the final y, I know the initial y, I know the initial velocity and know the acceleration. The only thing I don't know in that expression is time, okay? So I can solve for it. Let me just go ahead and do one quick thing. I know that the final y velocity is zero because this is the part that gets people tricked up. So but everything else is a number. That's a number, that's a number, that's a number. So how do you solve for t? People try to do some weird factoring and stuff like that. Don't try that, you're going to hurt yourself. Just jump on ahead, you have constant plus something times t plus something times t squared. That should say use the quadratic equation, okay? So I'm not going to give a quadratic equation lesson, I'm just going to go ahead and write down the solution for t. This is the a number, this is b and that's c, okay? So that means that t is going to be negative b, so that's our b, so negative v zero y plus or minus the square root of b squared minus four times a, which is, we're going to have a negative one-half, so it's going to be plus two. So four a c, so four a g, y zero, all of that over two a, so it's going to be, that's right there, negative, no, negative all of that over two a, so it's negative g. Is that right? I think that's right. Well let's just check, okay, I skipped some steps so it's good to check the units, let's check our units. These have to have the same units to add them together, so this is meter squared per second squared and here are meters per second squared times meters, so yes, I can add those together. When I take the square root I get meters per second squared, which allows me to add it to that, so that's good. And then I divide by meters per second squared divided by meters per second squared, no, this is meters per second divided by meters per second squared will give me seconds, so that doesn't mean I made it right, but it's a good sign that I didn't. I'm going to get two answers for when the ball gets back down to y equals zero, because math does not know what we're doing here. It says, okay, the ball was at y equals zero right here and right there, so we're going to get a negative time, okay. I don't want to finish the problem, because I don't want to get bogged down on all the numbers and take up a lot of time. You could put in your numbers here, you're going to get two times, one of them is going to be positive, use that one, and then take that time and put it up there and you get your final position, okay. Yes, it didn't work out the whole thing, but I think this will be good practice for you. It's not a terribly difficult problem, except that you have to use a quadratic equation and that throws people off. This is something I would expect you to be able to do, a problem like this.