Projectile motion (part 1)

Welcome back. I'm not going to do a bunch of projectile motion problems, and this is because I think you learn more just seeing someone do it, and thinking out loud, than all the formulas. I have a strange notion that I might have done more harm than good by confusing you with a lot of what I did in the last couple of videos, so hopefully I can undo any damage if I have done any, or even better-- hopefully, you did learn from those, and we'll just add to the learning. Let's start with a general problem. Let's say that I'm at the top of a cliff, and I jump-- instead of throwing something, I just jump off the cliff. We won't worry about my motion from side to side, but just assume that I go straight down. We could even think that someone just dropped me off of the top of the cliff. I know these are getting kind of morbid, but let's just assume that nothing bad happens to me. Let's say that at the top of the cliff, my initial velocity-- velocity initial-- is going to be 0, because I'm stationary before the person drops me or before I jump. At the bottom of the cliff my velocity is 100 meters per second. My question is, what is the height of this cliff? I think this is a good time to actually introduce the direction notion of velocity, to show you this scalar quantity. Let's assume up is positive, and down is negative. My velocity is actually 100 meters per second down-- I could have assumed the opposite. The final velocity is 100 meters per second down, and since we're saying that down is negative, and gravity is always pulling you down, we're going to say that our acceleration is equal to gravity, which is equal to minus 10 meters per second squared. I just wrote that ahead of times, because when we're dealing with anything of throwing or jumping or anything on this planet, we could just use this constant-- the actual number is 9.81, but I want to be able to do this without a calculator, so I'll just stick with minus 10 meters per second squared. It's pulling me down, so that's why the minus is there. My question is: I know my initial velocity, I know my final velocity, right before I hit the ground or right when I hit the ground, what's the distance? In this circumstance, what does distance represent? Distance would be the height of the cliff, and so how do we figure this out? What's the only formula that we know for distance, or actually the change in distance, but in this case, it's the same thing. Change in distance is equal to the average velocity. When you learned this in middle school, or probably even elementary school, you didn't say average velocity, because you always assumed velocity was constant-- the average and the instantaneous velocity was kind of the same thing. Now, since the velocity is changing, we're going to say the average velocity. So, the change in distance is equal to the average velocity times time. This should be intuitive to you at this point. Velocity really is just distance divided by time, or actually, change in distance divided by times change in time-- or, change in distance divided by change in times is velocity. Let me actually change this to change in time. Since we always assume-- or we normally assume-- that we start at distance is equal to 0, and we assume that start at time is equal to 0, we can write distance is equal to velocity average times time. Maybe later on we'll do situations where we're not starting at time 0, or distance 0, and in that case, we will have to be a little more formal and say change in distance is equal to average velocity the change in time. This is a formula we know, and let's see what we can figure out. Can we figure out the average velocity? The average velocity is just the average of the initial velocity and the final velocity. The average velocity is just equal to the average of these two numbers: so, minus 100 plus 0 over 2-- and I'm just averaging the numbers-- equals minus 50 meters per second. We were able to figure that out, so can we figure out time? We know also that velocity, or let's say the change in velocity, is equal to the final velocity minus the initial velocity. This is nothing fancy-- you don't have to memorize this. This hopefully is intuitive to you, that the change is just the final velocity minus the initial velocity, and that that equals acceleration times time. So what's the change in velocity in this situation? Final velocity is minus 100 meters per second, and then the initial velocity is 0, so the change in velocity is equal to minus 100 meters per second. I'm kind of jumping in and out of the units, but I think you get what I'm doing. That equals acceleration times time-- what's the acceleration? It's minus 10 meters per second squared, because I'm going straight down-- minus 10 meters per second squared times time. This is a pretty straightforward equation. Let's divide both sides by the acceleration, by the minus 10 meters per second squared, and you'll get time is equal to-- the negatives cancel out, as they should, because negative time is difficult, we're assuming positive time, and it's good we got a positive time answer-- but the negatives cancel out and we get time is equal to 10 seconds. There we have it: we figured out time, we figured out the average velocity, and so now we can figure out the height of the cliff. The distance is equal to the average velocity minus 50 meters per second times 10 seconds. The distance-- this is going to be an interesting notion to you-- the distance it's going to be minus 500 meters. This might not make a lot of sense to you-- what does minus 500 meters mean? This is actually right, because this formula is actually the change in distance. We said if we did it formally, it would be the change in distance. So if we have a cliff-- let me change colors with it-- and if we assume that we start at this point right here, and this distance is equal to 0, then the ground, if this cliff is 500 hundred meters high, your final distance-- this is the initial distance-- your final distance df is actually going to be at minus 500 hundred meters. We could have done it the other way around: we could have said this is plus 500 meters, and then this is 0, but all that matters is really the change in distance. We're saying from the top of the cliff to the ground, the change in distance is minus 500 meters. And minus, based on our convention, we said minus is down, so the change is 500 meters down, and that's height of the cliff. That's pretty interesting. If you go to a 500 meter cliff-- 500 is about 1,500 feet-- so that's roughly the size of maybe a very tall skyscraper, like the World Trade Center or the Sears Tower. If you jump off of something like that, assuming no air resistance, which is a big assumption, or if you were to drop a penny-- because a penny has very little air resistance-- if you were to drop a penny off of the top of Sears Tower or a building like that, at the bottom it will be going 100 meters per second. That's extremely fast, and that's why you shouldn't be doing it, because that is fast enough to kill somebody, and I don't want to give you any bad ideas if you're a bad person. It's just interesting that physics allows you to solve these types of problems. In the next presentation, I'm just going to keep doing problems, and hopefully you'll realize that everything really just boils down to average velocity-- change in velocity is acceleration times time, and change in distance is equal to change in time times average velocity, which we all did just now. I'll see you in the next presentation.