 Okay, let's see if I can do this in under 11 minutes. So here's the problem. A very interesting age old problem, height H, a rock right there, you know ancient man, you know when they were trying to learn about fire, this is what they did, they said hey, how do you, how do you find out how fast the rock is going when you drop it off a cliff? It's ancient. And it's on, it's okay. So how do I find out how fast it's going down at the bottom right before it hits? Right before it hits, because once it hits, well there's going to be another interaction involved, an interaction that I really can't account for. As it's falling, it's only interacting with the gravitational force from the earth, that's it. So I'm not sure what part of the curriculum you would insert this in. If you talk about kinematics and just saying things that fall will have an acceleration in the y direction of negative 9.8 meters per second squared, then you could just do that. If you wanted to talk about forces, I'd draw a force diagram. So we know the acceleration, we know the height, if I say v1y equals zero meters per second, then what's v2y? You notice I've not written this as a vector because we're only dealing with this in one dimension. Okay, so what do we know? I feel very influenced by what he just did, but I'm going to try to do it the way I would normally do it. So what, there are two really things we know. We know the definition of velocity, v average y is the change in y over the change in t, and we know that the definition of the y acceleration, which is constant in this case, so I don't have to say average, is change in vy over change in t. Con said something about air resistance. I don't think that's, if you're talking about this from a kinematics approach to saying things that fall have an acceleration of negative 9.8 meters per second squared, then you bypass the whole forcing anyway. So we don't need to talk about that right now. Okay, so I want to find the velocity at the bottom. So I can use this. This is v2y minus v1y over delta t. So I can solve this for v2y. This marker's not very good either. I'm going with it. v2y equals pi both sides about delta t and add v1y plus ay delta t. Okay, so yes, I know that this is zero, but I'll just leave it in there for now. I know that is a value, it's a negative value, but I'm going to leave the plus right there. But I don't know delta t. So up here I can solve for delta t. Delta t equals the change in y over the average velocity in the y direction. And average velocity, since the velocity is changing at a constant rate, then the average velocity, if I drew a graph of vy as a function of t, it would look like this. So if I want to find the average, this is vy, this is t. The average in this case is going to be just halfway, because it looks like a...it is. Okay, so v average y, maybe I say things twice too, is going to be v2y plus v1y over 2. So I can put that in for delta t. Yeah, I'm trying to find the final velocity. So v2y equals v1y plus ay, and then delta t is going to be delta y times 2 over v2y plus v1y. So that's going to be delta y times 2 over v2y plus v1y. Now one thing that I've done here is I didn't take into account anything about the initial conditions yet. So I've derived something more generic, really. That doesn't look right, but let's see. That's v, this is going to be meters per second squared times meters per second squared divided by meters per second. Okay, so it has the right units. Okay, so that's my general expression for v2y, but there's v2y down there also. So that's...we have to algebraically solve for that. So let's just do that real quick. I'm going to go ahead and say the initial y velocity is zero. So that means that I can multiply both sides by v2y, and I get v2y squared equals 2 delta y ay. And then I just take the square root of both sides, and I get the final velocity. Now so here I'm dealing with scalars, so the sign doesn't matter. A is the negative value, and delta y is what? Well, if I call this y equals zero, then y1 is h, and y2 is zero. So zero minus h is going to be delta y negative h. You don't have...there's no such thing as the origin. You can't see it. So you can put it wherever you want. I could put hy equals zero up here, too. In this case I'd have...the final y is going to be negative h, and the initial is zero. Either way, I get delta y is negative h. So I get this square root of 2, negative h, negative 9.8 meters per second squared. And you should check the units and everything like that. You should check as h gets larger, then this velocity gets larger. That makes sense, too. You get a plus or minus here. What does that mean? It really means that you don't know, because if you plot this...well, in terms of an energy standpoint, the solution algebraically to this, if you put in a positive or negative velocity, it should still work. Really, you know...the negative velocity would correspond to a negative time, or the positive velocity would. So as you're solving this equation, here's y as a function of t, and you're doing this, there's the motion, and you're finding the velocity right there. But at that height, also, there's a velocity over here that's negative. So it gives you two values, one's a negative time. Okay, I hope that didn't take more than 11 minutes. Eight. Awesome.