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Current time:0:00Total duration:8:57

Video transcript

well now use that that equation we just derived to to go back and solve a or at least address that same problem we were doing before so let's write that equation down again so actually let's write the problem down let's say I have the cliff again and so my initial distance is zero but it goes down five hundred meters I'm not going to redraw the cliff because it takes a lot of space up on my my limited chalkboard so we know that the change in distance change in distance is equal to minus 500 meters and I'm going to still use the example where I don't just drop the ball or the penny or whatever I'm throwing off the cliff I actually throw it straight up so it's going to go up and slow down from gravity and then you know go to zero velocity then start decelerating downwards or you can even say decelerating in the other direction but the initial velocity VI is equal to let's say 30 meters per second and of course we know that the acceleration is equal to minus ten meters per second squared because we're acceleration gravity is always pulling downwards or towards the center of our planet if we wanted to figure out the final velocity we could have just used the formula we did this in the last video VF squared is equal to VI squared plus two ad but now what I want to do is see if we I'm going to use the the formula we learned in the very last video to figure out how long does it take to get to the bottom of to hit the ground so let's use that formula we that we derived and I'll write in a different color we derived that the change in distance change in distance is equal to the initial velocity times time plus acceleration times squared over two that's initial velocity so the change in distance is minus 500 minus 500 is equal to the initial velocity well that's positive going upwards 30 meters per second 30 T I'm not going to write the unit's right now because it'll just make things I'll run out of space but you can redo it with the units and see that the unit's do work out when you square time you have to square the time units although we're solving for time plus acceleration plus acceleration acceleration is minus ten we're going to divide it by two right so it's minus five minus 5t squared so we have minus 500 is equal to 30 T and we could you know plus minus 5t squared we just say minus 5t squared to get rid of this plus so at first you're saying well Sal these there's t that's you know T to the first to the second how do I solve this and and hopefully you've taken algebra 2 or Algebra one in some places and you know you remember how to solve this otherwise you're about to learn the quadratic equation although I recommend you go back and you learn about factoring in the quadratic equation which there are videos on that I've put on on YouTube so I hope you watch those first if you don't remember but we can do a list but let's put these two right terms on the left hand side and then we'll use a quadratic equation to solve it sound think this is easy to factor so we get 5 T squared minus 30 t minus 500 is equal to zero I just took these terms and put them on the left side and we could divide both sides by 5 just to I don't know just to simplify things so we get T squared minus 6t minus 100 is equal to 0 right and I could do that because I 0 divided by 5 is just 5 so it just cleaned it up a little bit so let's use a quadratic equation and for those of us who need a refresher I'll write it down so the roots of any quadratic and in this case you know it's T we're solving for T will equal negative B plus or minus the square root of b squared minus 4ac over 2a where a is the coefficient on this term B is the coefficient on this term negative 6 and C is the constant so minus 100 so let's just solve so we get T is equal to negative B so negative this term well this term is negative 6 so if we make it a negative it becomes plus 6 so becomes 6 plus or minus the square root of B squared so it's minus 6 squared 36 minus 4 minus 4 times a the coefficient on a is here that's just 1 times 1 for AC C is a constant term minus 100 minus 4 times 1 times minus 100 times minus 100 all of that over 2a well a is 1 again so all of that over 2 that just equals 6 plus or minus the square root this is minus 4 times minus 100 so these become pluses so it becomes 36 plus 400 right so 6 plus or minus 436 divided by 2 and this is not a clean number and if you type it into your calculator it's something on the order of about 20 20.9 we can just say approximately 21 you might want to get the exact number if you were actually doing this on a test or trying to send something to Mars but for our purposes I think you'll get the point so I'll say it approximately now because we're going to be a little off but just have clean numbers this is approximately 21 it's a little it's like 20 point 9 we'll say 6 plus or minus now let me just write twenty point nine twenty point nine over two so T is approximately six so let me ask you a question if I do if I do so let's do if I do six minus twenty point nine what do I get I get a negative number and does a negative time make sense no it does not that means that somehow in the in the past well I don't want to get philosophical the negative time in this context will not make sense so really we can just stick to the plus right because six minus 20 is negative so six will just there's only one time that solves this in a meaningful way so time is approximately equal to six plus twenty point nine so that's twenty six point nine over 2 and that equals what thirteen point four five 13 0.45 and seconds time is equal to that's interesting I think if you remember way back may 4 or 5 videos ago when we first did this problem we just dropped we just dropped the penny straight from from the height and I've actually in that problem I gave you the time I said it took 10 seconds to hit the ground and we worked backwards to figure out that the cliff was 500 meters high now so if you if it you're at the top of a 500 meter cliff or building and you drop something that has no air resistance like a penny that has very little air resistance it would take 10 seconds to reach the ground assuming you know all of our assumptions about gravity whatever but if you were to throw the penny straight up you know off the edge of the cliff at 30 meters per second right here it's going to take 13.5 roughly 13 and a half seconds to reach the ground so it takes a little bit longer and that that should make sense because actually let me I have time to draw a little picture in the first case Oh me in the first case I just took the penny and it's motion just went straight down in the second case I took the penny it first went up and then it went down right so it had all the time when it went up and then it had it kind of went down a longer distance so it makes sense this time is 10 10 seconds well this time was thirteen point four five seconds so you can kind of say that it took it took well you actually can't say that well you well yeah I don't want to I don't get too involved but hopefully this makes sense to you if you've got a smaller number here you should you would have gone and checked your work because why would it take less time when I throw the the object straight up so hopefully that that gave you a little bit of more intuition and you really do have all in your arsenal now all of the equations and hopefully the intuition you need to to solve basic projectile problems I'll now do a couple of more videos where I just do a bunch of problems just to really really drive the point home and then I'll expand these problems to two to demit two dimensions and angles and and before we get there you might want to refresh your trigonometry I'll see you soon