In the last video, I dropped a penny from the top of a cliff and it started off at zero velocity. I was thinking it was stationary and that the bottom was a 100m/s. We used that to figure out how high the cliff was and we figured out that the cliff was 500m high. So, what I want to do now is do that same problem but let’s do it in a general form. Let’s see if we can figure out kind of a general formula for a problem like that.
So, let’s say that you have the same thing. Let’s say the initial velocity. Well, actually we’re going to keep everything. Let’s say you’re given the initial velocity. You’re given the final velocity. You’re given the acceleration and you want to figure out the distance.
So, doing it the exact same way we did in that last presentation but when I was—about formulas, we know that the change in distance is equal to the average velocity times time. We could actually say the change in time but I’ll just say it with time because we always assume that we start time = 0. Well, we know that the average velocity is the final velocity plus the initial velocity divided by two, so that’s the average velocity. This is the same thing as this and that times time.
Well, what’s the time? Well, you could figure out the time by saying, “Well, we know how fast we’re accelerating and we know the initial and final velocity, so we can figure out how long we had to accelerate that fast to get that change in velocity.” Well, another way of saying that, probably the simpler way of saying that is change in velocity which is the same thing as the Vf - Vi = a × t. Or if you want to solve for time, we could say the time—if I just divide both sides of this equation by a, t = Vf - Vi/a. So then, we can take that and substitute that into this equation. And remember, this is all change in distance. So, you say change in distance is equal to, (Vf + Vi/2)(Vf - Vi/a). And then if we do a little multiplying of expressions on the top, you might have recognized this, this would be Vf2 - Vi2 and then you multiply the denominators over 2a. So, the change in distance is equal to Vf2 - Vi2 over 2a. So, the change in distance is equal to Vf2 - Vi2 ÷ 2 × a.
Now, we could play around with this a little bit. And if we assume that we start at distance = 0, we could just write d here and that might simplify things. If we multiply both sides by 2a, we get 2 and I am just going to switch this to distance. If we assume that we always start at distance = 0, di or initial distance is always at 0.0 but we could write 2ad. I am just multiplying both sides by 2a = Vf2 - Vi2 or you could write it as Vf2 = Vi2 + 2ad. I don’t know what your physics teacher might show you or what’s written in your Physics book but one of these variations will show up in your Physics book. But the reason why I wanted to show that previous problem first is I wanted to show you that you could actually figure out these problems without having to always memorize formulas and resort to the formula but with that said, it’s probably not a bad idea to memorize some formulas, although you should understand how it was derived and when to apply it.
So, now that you have memorized it or I’ve shown you that maybe you don’t have to memorize it, let’s use this. So, let’s say I have the same cliff and it was 500m high. And this time with the penny, instead of just dropping it straight down, I am going to throw it straight up at positive 30m/s. The positive matters because remember we said negative is down, positive is up, that’s just the convention we use.
So, let’s use this formula, any of versions of this formula to figure out what our final velocity was when we hit the bottom of the ground. Well, this is probably the easiest formula to use because it actually solves for final velocity. So, we can say that the final velocity Vf2 = Vi2. So, what’s our initial velocity? It is +30m/s. So, it’s 30m/s2 + 2ad. So, 2a is the acceleration of gravity which is -10 because it’s going down. So, it’s 2 × -10 and then what’s the height? What’s the change in distance? Actually, I should be correct about using change in distance because it matters for this problem. So in this case, the final distance is equal to -500 and the initial distance is equal to zero. So, the change in distance is -500. So, what do we get? So, we get Vf2 = 900. And then the negative cancels out. 10 × 500 is 5,000. 5,000 × 2 is 10,000. So, Vf2 = 10,900. So, the final velocity is equal to the √10,900. So, what is that? So, it’s about a 104m/s. So, my final velocity is approximately 104m/s.
If I just drop it straight from the top, we figure out in the last problem at the end, I’m in a 100m/s but this time, if I throw it straight up at 30m/s, when the penny hits the ground, it’s actually going even faster. And so, you might want to think about why that is. And you might realize it because when I throw it up at 30m/s, the highest point of the penny is going to be higher than 500m. It’s going to make some positive distance first and then it’s going to come down, so it’s going to have even more time to accelerate. I think that makes some intuitive sense to you.
So, that’s all the time I have now. In the next presentation, maybe I’ll use this formula to solve a couple of other types of problems.
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