 Okay, so we're still talking about, I'm still talking about the momentum principle. I did those two examples with the rockets on it. Let me do an example of something that we've kind of already seen. So suppose I have, and this is a great example because you can do it a whole bunch of different ways. So let's just say I have a ball to make it a simple case. Here's a ball right there, and it starts off a high, let's just say a height h, we can put in a number later if we want, above the ground, and it starts with an initial velocity of zero meters per second, and I drop it. And I guess I want to find out two things and try to think what I could find out. I could find out how fast it's going when it gets to the bottom, and I could find out how long it took. And then you can see we can do all sorts of projectile problems using momentum principle. Okay, so with momentum principle, the first thing we want to think about is what forces are acting on the object because we need the net force. So here we have, if I let go of the ball, then I only have one thing acting on it, and that's the gravitational force. And near the surface of the earth, I can call that as mg, where g is the best word for it's the gravitational field. Okay, so that's my net force. So let me write it down in the momentum principle, f net equals the change in momentum over the change in time. Okay, and what force do I have on there? It's just mg. So let me now, this is clearly just a one-dimensional problem, x, y, so let me write this in the y direction. In the y direction what's the net force? It's negative mg, right, because g is now, that's the vector g which is down, but in the y direction that'd be a negative sign, so negative mg. Now the change in momentum is going to be, I'll just write it out as mv2y minus mv1y over delta t. The initial velocity was zero, I'll just go ahead and remember that. And here, let me go ahead and actually let me write this, let me say that's not zero because in the generic case I could write this as m delta v over delta t, and here you see something extremely important. The gravitational force is proportional to the mass, and since the change in momentum is proportional to the mass, or the change in velocity is proportional to the mass in this case, then the mass cancels, and this gives you something really, so I get negative g equals delta v over delta t, or the change in velocity, the acceleration, is equal to negative g, this constant. So if I take a heavy object and a light object and I drop them, they're going to have the same acceleration if gravity is the only thing acting on them. So essentially the more mass of something is the greater the gravitational force, but also the harder it is to speed it up, or to change its momentum, and since these masses are the same, this is technically the gravitational mass, and this is the called the inertial mass, they happen to be the same thing, okay, they cancel, so things follow the same acceleration. Okay, back to this problem though. So my initial velocity is zero, so I get v2y, if I multiply by delta t equals negative g delta t. Okay, but I don't know the time, right, but this is just a definition of the acceleration and the change in velocity, but I don't know the time. Okay, so how can I get that? Well, here I can use, go back to average velocity, so the average velocity in the y direction is going to be v2y plus v1y over 2, and that initial velocity was zero, so this is v2y over 2. So that's the average velocity that falls down here, so then I can say that's also equal to delta y over delta t, because the average velocity is delta y over delta t, so v2y over 2 equals delta y over delta t, and what's delta y? If this is y equals zero and that's y1, then it starts off at h and it ends up at zero, so the final is zero minus the initial of h, so this is going to be negative h over delta t. Okay, let me find delta t, that's what I really want. So if I multiply both sides by delta t and divide by, or multiply by 2 over v2, I get delta t equals negative 2h, yeah, that's fine. Okay, over v2y. Okay, so why did I pause for a minute there? I paused because I thought, wait, I'm going to have a negative time, and I'm not going to have a negative time. The y velocity is negative, the y component of velocity is the negative number two, so it's going to be positive, but I can check the units, meters, meters per second does give me seconds. Also, I can check other things. What if I drop it from a higher height, it's going to take longer? Okay, so I don't know v2, I can't get, even if I knew h was one meter, I couldn't do the problem, but I can, now I have this equation and this equation. I can solve, I have two things I don't know. I don't know v2y and I don't know t. So I can use some algebra to find out what I don't know. What one do I want to solve for? First, I'm going to solve for v2y, I shouldn't read it, so v2y is negative g delta t, which is that negative 2h over v2y. Those cancel. Multiply both sides by v2y, v2y squared equals 2gh. v2y equals the square root of 2gh. Does that have the right units? 1g is in Newtons per kilogram, which is the same as meters per second squared. Times meters is meters squared per second squared. Take the square root, I do get the right units. And then going back over here, I can get delta t. And this is actually plus, this is going to give me a plus or minus because, and we know it has to be negative because it's going down. So I'm going to take the negative version of that over here to make this thing work. Then over here I get delta t equals negative h over negative the square root of 2gh. I can rewrite this as, if I have 2 divided by the square root of 2, that's the square root of 2 on the top, and h divided by the square root of h is the square root of h on the top. So I get the square root of 2h over g. Meters over meters per second squared gives me second squared, take the square root, and I do get the correct time. Okay, so we did this problem before with kinematics, just using kinematic equations, and we kind of did that here too, if you see. But I started with momentum principle. Okay, the most important thing to see is why do things fall and have an acceleration of negative 9.8 meters per second squared. You could say it's because of the momentum principle.