 Okay, so hopefully you can hear me and hopefully you can see everything but we'll just try this out So projectile motion Let's look at projectile motion in projectile motion. The key thing is We define projectile motion to be An object moving only under the influence of gravity. So let's just let's just start Okay, so suppose I take a ball and I shoot it Horizontally while it's in its path right here. What force are acting on it? Don't say it. Don't say the force from the cannon that shot it. No, it's not It's not acting on it anymore. The only force acting on it is gravity, which is straight down. I'll write that as MG Can you see that? Okay? That's a little small. Well, we'll just try it out. Okay so From that we can write down Newton's second law or the momentum principle, however you want to call it. I'll write that as F that Equals ma and so in two dimensions. I can write f net x equals ma x F net y equals ma y so The only force is in the y direction. There's no force in the x-direction. So this equation says f net in the x-direction Zero, so that's zero. So a x equals zero So in the x-direction, I have the following equation can completely describe the motion of The particle in the x-direction x final equals x initial plus v x Assuming time starts at zero Okay, because there's no acceleration. This is just the definition of average velocity is change in position or change in time or written in a different way So that's that Then nothing else to say in the x-direction in the y-direction. I have the net force is going to be in negative MG Where G is positive 9.8 Newton's per kilogram So I can solve this for a y equals negative G sets acceleration in the y-direction So now I have a constant acceleration. I can use the kinematic equations I'm just gonna write down one of them. You can go back and look at kinematic equations But this is the important so in the y-direction. I have y final equals y initial plus v y initial times T minus one half G T squared just your plain kinematic equation free fall the key is this motion is independent of this motion except for The time it takes so it's just like a ball moving at a constant speed up there and a ball falling straight down down to there And the only thing they have in common is the time the time it takes for this constant horizontal motion is equal to The time it takes of all the falls straight down So that's simple case This is always gonna be true. Okay, the only thing that you need to worry about is And then so what would you do what it depends on what you're trying to do Suppose you're trying to solve for how far it goes horizontal. That's a common case Well, here if I shoot this ball horizontally, I can call the floor down here Y equals zero meters So in that case the final y would be zero Initial y would be let's call it H What's the initial y velocity if it's shot that way? It's not moving at all in the y direction. So the initial y velocity is zero. So that term is gone I'm just left with this one minus one half g t squared now I can solve for the time and see this is the trick the y direction I can use to solve for the time and then use that same time in the x direction. So here if I Let's say I subtract h from both sides and I get negative h equals negative one half g t squared and Then I can divide both sides by negative g over two and I get T squared equals two h over g T equals the square root of two h Okay, and then I can if I want to find out how far it went and if I know the initial velocity in this case The initial velocity is all in the x direction So I can solve for let's say it starts at x equals zero x final It's just going to be vx, which is the launch velocity times t, which is this Okay. Now one more quick case. What if I change this? So that I launch it at an angle with some velocity v zero at an angle theta above the horizon So what's going to change? Well, what's going to change is now my initial x velocity is not v not Am I initial y velocity not zero? So I can write vx initial. It's going to be The zero the x the y So there you have a right triangle So using your normal trig rules vx is going to be v zero cosine theta V y zero which is just the initial velocity. It's going to be v zero sine theta And then over here I could do if I want to find out how far it went Now this would not be zero my initial y velocity is that value So this would change and it'd be a little bit more difficult to solve In fact, if you try to do your normal algebraic rules you'll fail You need to use the quadratic equation You can't remember what that is look it up But you can still find the time and plug it into here and not all not every single can project on motion problems It's going to be the same way, but the key thing is Projectile motion you can treat the x and the y motion separately Except they have the same time In the x direction the acceleration zero in the y direction the accelerations negative g So I went kind of fast because I don't want to make a an hour recording and you can pause it and you can replay it So let me know if this helps