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Current time:0:00Total duration:9:15

Video transcript

in the last video I dropped myself or penny from the top of a cliff and you know start off at zero velocity because it was stationary and at the bottom was 100 meters per second and we use that to figure out how high the cliff was right and we got we figured out that the cliff was 500 meters high so what I want to do now is is let's do that same problem but let's do it in the general form is if we can figure out kind of a general formula for 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 this is what you're given and you want to figure out the distance so doing it the exact same way we did in that last in that last presentation but when our just going skating formulas we know that the change in distance the change in distance is equal to the average velocity times unless we could actually say the change in time but I'll just say it with time because we always assume that we start at time equals zero times time well we know that the average velocity the average velocity is what it's the final velocity plus the initial velocity divided by two so that's the average velocity so let me let me highlight in the right this is the same thing as this and then that times time well what's the time well the time you can 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 much how long we had to accelerate that fast to get that change in velocity or another way of saying that probably a simpler way of saying that is change in velocity change in velocity which is the same thing as the final velocity minus the initial velocity is equal to acceleration times time right or if you want to solve for time you could say the time all right if I just divide both sides of this equation by a is equal to VF minus VI divided by a right so then we could 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 so the first this term let me write this term in in yellow VF plus V I over to let me write this term and I don't know green so that's x VF minus VI over a I like the screen I don't know why I don't never use it and then if we multiple you know multiplying of expressions on the top you might recognize this this would be VF squared minus VI squared and then you multiply the denominators over to a so the change in distance is equal to VF squared minus VI squared over 2a that's exciting let me write that over again image invert colors so the change in distance the change in distance is equal to the VF squared minus VI squared divided by the two times acceleration and we could play around with this a little bit and if we assume that we start at distance is equal to zero we could just write D here and that might simplify things if we multiply both sides by 2a we get two and I'm just going to switch this to distance if we if we assume that we always start at distance is equal to zero d di or initial distance is always at point zero but we could write to ad right I'm just multiplying both sides by 2a is equal to V F squared minus VI squared or you could write it as VF squared is equal to VI squared plus 2 ad and you'll I don't know what your physics teacher might show you or what's written in your physics book but one very one of these variations will show up in your physics book but the reason why I wanted to show you that previous problem first is that I wanted to show you that you could actually figure out these problems without having to always you know memorize formulas and resort to the formula but with that set is probably not a bad idea to memorize some form of this formula although you should understand how it was derived and and when to apply it so let's let's what that's you know now that you have memorized it or I've shown you that maybe you don't have to memorize it let's let's use this so let's say I have the same cliff and it has now turned purple and it's what did I say it was it was 500 meters high right it's a 500 meter high cliff and this time with the penny instead of just dropping it straight down I'm going to throw it straight up at I don't know let's say positive 30 meters per second and and the positive matters because remember we said negative is down positive is up that's just the convention we use so let's use let's use this formula which you know any any very version of this formula to figure out let's use it to figure out what our final velocity was when we when we hit when we hit to the bottom of the ground well this is this is probably the easiest formula to use because it actually solves for final velocity so we can say that the final velocity VF squared is equal to the initial velocity squared so what's our initial velocity it's plus 30 meters per second right so it's 30 meters per second squared +2 ad so 2a is the acceleration of gravity which is -10 because it's going down right so it's 2 I want to run on space times minus 10 I'm going to give up the unit's for a second just so I don't run out of space 2 times minus 10 and then what's the height what's the change in distance actually I should be I should be correct about using change in distance because it matters for this problem right so in this case this is this is the final distance is equal to minus 500 right and the initial distance is equal to 0 so the change in distance is minus 500 right so what does this get us so we get VF squared is equal to 900 and then the negatives cancel out 10 times 500 is five thousand five thousand times to five thousand five to two is ten thousand so VF squared VF squared is equal to ten thousand nine hundred so the final velocity is equal to the square root of ten thousand nine hundred so what is that well let me let me let me let me bring over my trusty windows provide a default calculator let's see what is it ten thousand nine hundred square root so it's about a hundred four meters per second one hundred four meters per second so my let me get this out of the way so my final velocity is approximately that squiggly equals is approximately 104 meters per second that's interesting if I just drop something if I just drop it straight from the top we figured out the last problem at the end I'm at a hundred meters per second right but this time if I throw it straight up at 30 meters per second when I hit that when the penny hits the ground it's actually going even faster so you might want to think about why that is and and you might realize it because when I throw it up the highest point of the penny if I throw it up at 30 meters per second the highest point of the penny is going to be higher than 500 meters right 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 and you can I mean I think that makes some intuitive sense to you so I that's all the time I have now and in the next presentation maybe I'll use this formula to solve a couple of other types of problems