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Rolling without slipping problems (video) | Khan Academy

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Video transcript

- [Voiceover] All the problems we've been dealing with so far have essentially been happening in one dimension. You could go forward or back. So you could go forward or back. Or right or left. Or you could go up or down. What I wanna start to talk about in this video is what happens when we extend that to two dimensions or we can even just extend what we're doing in this video to three or four, really an arbitrary number of dimensions. Although if you're dealing with classical mechanics you normally don't have to go more than three dimensions. And if you're gonna deal with more than one dimension, especially in two dimensions, we're also gonna be dealing with two-dimensional vectors. And I just wanna make sure, through this video, that we understand at least the basics of two-dimensional vectors. Remember, a vector is something that has both magnitude and direction. So the first thing I wanna do is just give you a visual understanding of how vectors in two dimensions would add. So let's say I have a vector right here. That is vector A. So, once again, its magnitude is specified by the length of this arrow. And its direction is specified by the direction of the arrow. So it's going in that direction. Now let's say I have another vector. Let's call it vector B. Let's call it vector B. It looks like this. Now what I wanna do in this video is think about what happens when I add vector A to vector B. So there's a couple things to think about when you visually depict vectors. The important thing is, for example, for vector A, that you get the length right and you get the direction right. Where you actually draw it doesn't matter. So this could be vector A. This could also be vector A. Notice, it has the same length and it has the same direction. This is also vector A. I could draw vector A up here. It does not matter. I could draw vector A up there. I could draw vector B. I could draw vector B over here. It's still vector B. It still has the same magnitude and direction. Notice, we're not saying that its tail has to start at the same place that vector A's tail starts at. I could draw vector B over here. So I can always have the same vector but I can shift it around. So I can move it up there. As long as it has the same magnitude, the same length, and the same direction. And the whole reason I'm doing that is because the way to visually add vectors... If I wanted to add vector A plus vector B... And I'll show you how to do it more analytically in a future video. I can literally draw vector A. I draw vector A. So that's vector A, right over there. And then I can draw vector B, but I put the tail of vector B to the head of vector A. So I shift vector B over so its tail is right at the head of vector A. And then vector B would look something like this. It would look something like this. And then if you go from the tail of A all the way to the head of B, all the way to the head of B, and you call that vector C, that is the sum of A and B. And it should make sense, if you think about it. Let's say these were displacement vectors. So A shows that you're being displaced this much in this direction. B shows that you're being displaced this much in this direction. So the length of B in that direction. And if I were to say you have a displacement of A, and then you have a displacement of B, what is your total displacement? So you would have had to be, I guess, shifted this far in this direction, and then you would be shifted this far in this direction. So the net amount that you've been shifted is this far in that direction. So that's why this would be the sum of those. Now we can use that same idea to break down any vector in two dimensions into, we could say, into its components. And I'll give you a better sense of what that means in a second. So if I have vector A. Let me pick a new letter. Let's call this vector "vector X." Let's call this "vector X." I can say that vector X is going to be the sum of this vector right here in green and this vector right here in red. Notice, X starts at the tail of the green vector and goes all the way to the head of the magenta vector. And the magenta vector starts at the head of the green vector and then finishes, I guess, well where it finishes is where vector X finishes. And the reason why I do this... And, you know, hopefully from this comparable explanation right here, says, okay, look, the green vector plus the magenta vector gives us this X vector. That should make sense. I put the head of the green vector to the tail of this magenta vector right over here. But the whole reason why I did this is, if I can express X as a sum of these two vectors, it then breaks down X into its vertical component and its horizontal component. So I could call this the horizontal component, or I should say the vertical component. X vertical. And then I could call this over here the X horizontal. Or another way I could draw it, I could shift this X vertical over. Remember, it doesn't matter where I draw it, as long as it has the same magnitude and direction. And I could draw it like this. X vertical. And so what you see is is that you could express this vector X... Let me do it in the same colors. You can express this vector X as the sum of its horizontal and its vertical components. As the sum of its horizontal and its vertical components. Now we're gonna see over and over again that this is super powerful because what it can do is it can turn a two-dimensional problem into two separate one-dimensional problems, one acting in a horizontal direction, one acting in a vertical direction. Now let's do it a little bit more mathematical. I've just been telling you about length and all of that. But let's actually break down... Let me just show you what this means, to break down the components of a vector. So let's say that I have a vector that looks like this. Let me do my best to... Let's say I have a vector that looks like this. It's length is five. So let me call this vector A. So vector A's length is equal to five. And let's say that its direction... We're gonna give its direction by the angle between the direction its pointing in and the positive X axis. So maybe I'll draw an axis over here. So let's say that this right over here is the positive Y axis going in the vertical direction. This right over here is the positive X axis going in the horizontal direction. And to specify this vector's direction I will give this angle right over here. And I'm gonna give a very peculiar angle, but I picked this for a specific reason, just so things work out neatly in the end. And I'm gonna give it in degrees. It's 36.8699 degrees. So I'm picking that particular number for a particular reason. Now what I wanna do is I wanna figure out this vector's horizontal and vertical component. So I wanna break it down into something that's going straight up or down and something that's going straight right or left. So how do I do this? Well, one, I could just draw them, visually, see what they look like. So its vertical component would look like this. It would start... Its vertical component would look like this. And its horizontal component would look like this. Its horizontal component would look like this. The horizontal component, the way I drew it, it would start where vector A starts and go as far in the X direction as vector A's tip, but only in the X direction, and then you need to, to get back to the head of vector A, you need to have its vertical component. And we can sometimes call this, we could call the vertical component over here A sub Y, just so that it's moving in the Y direction. And we can call this horizontal component A sub X. Now what I wanna do is I wanna figure out the magnitude of A sub Y and A sub X. So how do we do that? Well, the way we drew this, I've essentially set up a right triangle for us. This is a right triangle. We know the length of this triangle, or the length of this side, or the length of the hypotenuse. That's going to be the magnitude of vector A. And so the magnitude of vector A is equal to five. We already knew that up here. So how do we figure out the sides? Well, we could use a little bit of basic trigonometry. If we know the angle, and we know the hypotenuse, how do we figure out the opposite side to the angle? So this right here, this right here is the opposite side to the angle. And if we forgot some of our basic trigonometry we can relearn it right now. Soh-cah-toa. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we have the angle, we want the opposite, and we have the hypotenuse. So we could say that the sine of our angle, the sine of 36.899 degrees, is going to be equal to the opposite over the hypotenuse. The opposite side of the angle is the magnitude of our Y component. ...is going to be equal to the magnitude of our Y component, the magnitude of our Y component, over the magnitude of the hypotenuse, over this length over here, which we know is going to be equal to five. Or if you multiply both sides by five, you get five sine of 36.899 degrees, is equal to the magnitude of the vertical component of our vector A. Now before I take out the calculator and figure out what this is, let me do the same thing for the horizontal component. Over here we know this side is adjacent to the angle. And we know the hypotenuse. And so cosine deals with adjacent and hypotenuse. So we know that the cosine of 36.899 degrees is equal to... Cosine is adjacent over hypotenuse. So it's equal to the magnitude of our X component over the hypotenuse. The hypotenuse here has... Or the magnitude of the hypotenuse, I should say, which has a length of five. Once again, we multiply both sides by five, and we get five times the cosine of 36.899 degrees is equal to the magnitude of our X component. So let's figure out what these are. Let me get the calculator out. Let me get my trusty TI-85 out. I wanna make sure it's in degree mode. So let me check. Yep, we're in degree mode right over there. Don't wanna... Make sure we're not in radian mode. Now let's exit that. And we have the vertical component is equal to five times the sine of 36.899 degrees, which is, if we round it, right at about three. So this is equal to... So the magnitude of our vertical component is equal to three. And then let's do the same thing for our horizontal component. So now we have five times the cosine of 36.899 degrees, is, if once again we round it to, I guess, our hundredths place, we get it to being four. So we get it to being four. So we see here is a situation where we have... This is a classic three-four-five Pythagorean triangle. The magnitude of our horizontal component is four. The magnitude of our vertical component, right over here, is equal to three. And once again, you might say, Sal, why are we going through all of this trouble? And we'll see in the next video that if we say something has a velocity, in this direction, of five meters per second, we could break that down into two component velocities. We could say that that's going in the upwards direction at three meters per second, and it's also going to the right in the horizontal direction at four meters per second. And it allows us to break up the problem into two simpler problems, into two one-dimensional problems, instead of a bigger two-dimensional one.

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Visualizing vectors in 2 dimensions (video) | Khan Academy

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Voiceover] All the problems we've been dealing with so far have essentially been happening in one dimension. You could go forward or back. So you could go forward or back. Or right or left. Or you could go up or down.

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You are by now probably used to the idea of measuring angles in degrees. We use it in everyday language. We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle. And we'd often use this symbol, just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you have essentially done one full rotation. And if you watch figure skating on the Olympics, and someone does a rotation, they'll say, oh, they did a 360. Or especially in some skateboarding competitions, and things like that. But the one thing to realize-- and it might not be obvious right from the get-go-- but this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. And if you think about it, you'll say, well, why do we call a full rotation 360 degrees? And there's some possible theories. And I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars. And even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1/360 of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot. And they had a base 60 number system. So they had 60 symbols. We only have 10. We have a base 10. They had 60. So in our system, we like to divide things into 10. They probably liked to divide things into 60. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. And that alternate way-- even though it might not seem as intuitive to you from the get-go-- in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena. But they might use what we're going to define as a radian. There's a certain degree of purity here-- radians. So let's just cut to the chase and define what a radian is. So let me draw a circle here, my best attempt at drawing a circle. Not bad. And let me draw the center of the circle. And now let me draw this radius. And let's say that this radius-- and you might already notice the word radius is very close to the word radians. And that's not a coincidence. So let's say that this circle has a radius of length r. Now let's construct an angle. I'll call that angle theta. So let's construct an angle theta. So let's call this angle right over here theta. And let's just say, for the sake of argument, that this angle is just the exact right measure so that if you look at the arc that subtends this angle-- and that seems like a very fancy word. But let me draw the angle. So if you were to draw the angle-- so if you look at the arc that subtends the angle, that's a fancy word. That's really just talking about the arc along the circle that intersects the two sides of the angles. So this arc right over here subtends the angle theta. So let me write that down. Subtends this arc, subtends angle theta. Let's say theta is the exact right size so that this arc is also the same length as the radius of the circle. So this arc is also of length r. So given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? Well, the most obvious one, if you kind of view a radian as another way of saying radiuses, or I guess radii. Well, you say, look, this is subtended by an arc of one radius. So why don't we call this right over here one radian, which is exactly how a radian is defined. When you have a circle, and you have an angle of one radian, the arc that subtends it is exactly one radius long. Which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degrees, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly the arc length that is subtending the angles. So let's do a couple of thought experiments here. So given that, what would be the angle in radians if we were to go-- so let me draw another circle here. So that's the center, and we'll start right over there. So what would happen if I had an angle-- if I wanted to measure in radians, what angle would this be in radians? And you could almost think of it as radiuses. So what would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? Well, let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle. Well, what's the circumference of a circle in terms of radiuses? So if this has length r, if the radius is length r, what's the circumference of the circle in terms of r? Well, we know that. That's going to be 2 pi r. So going back to this angle, the length of the arc that subtends this angle is how many radiuses? Well, it's 2 pi radiuses. It's 2 pi times r. So this angle right over here, I'll call this a different-- well, let's call this angle x. x in this case is going to be 2 pi radians. And it is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. So given that, let's start to think about how we can convert between radians and degrees, and vice versa. If I were to have-- and we can just follow up over here. If we do one full revolution-- that is, 2 pi radians-- how many degrees is this going to be equal to? Well, we already know this. A full revolution in degrees is 360 degrees. Well, I could either write out the word degrees, or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're kind of using units in both cases. Now, if we wanted to simplify this a little bit, we could divide both sides by 2. In which case, on the left-hand side, we would get pi radians would be equal to how many degrees? Well, it would be equal to 180 degrees. And I could write it that way, or I could write it that way. And you see over here, this is 180 degrees. And you also see if you were to draw a circle around here, we've gone halfway around the circle. So the arc length, or the arc that subtends the angle, is half the circumference. Half the circumference is pi radiuses. So we call this pi radians. Pi radians is 180 degrees. And from this, we can come up with conversions. So one radian would be how many degrees? Well, to do that, we would just have to divide both sides by pi. And on the left-hand side, you'd be left with 1-- I'll just write it singular now. 1 radian is equal to-- I'm just dividing both sides. Let me make it clear what I'm doing here, just to show you this isn't some voodoo. So I'm just dividing both sides by pi here. On the left-hand side, you're left with 1. And on the right-hand side, you're left with 180/pi degrees. So 1 radian is equal to 180/pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way. If I were to have 1 degree, how many radians is that? Well, let's start off with-- let me rewrite this thing over here. We said pi radians is equal to 180 degrees. So now we want to think about 1 degree. So let's solve for 1 degree. 1 degree, we can divide both sides by 180. We are left with pi/180 radians is equal to 1 degree. So pi/180 radians is equal to 1 degree. This might seem confusing and daunting. And it was for me the first time I was exposed to this, especially because we're not exposed to this in our everyday life. But what we're going to see over the next few examples is that as long as we keep in mind this whole idea that 2 pi radians is equal to 360 degrees, or that pi radians is equal to 180 degrees, which is the two things that I do keep in my mind. We can always re-derive these two things. You might say, hey, how do I remember if it's pi/180 or 180/pi to convert the two things? Well, just remember, which is hopefully intuitive, that 2 pi radians is equal to 360 degrees. And we'll work through a bunch of examples in the next video, to just make sure that we're used to converting one way or the other.

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Intro to radians (video) | Trigonometry | Khan Academy

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You are by now probably used to the idea of measuring angles in degrees. We use it in everyday language. We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle.

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- [Instructor] Let's talk about how to handle a horizontally launched projectile problem. These, technically speaking, if you already know how to do projectile problems, there is nothing new, except that there's one aspect of these problems that people get stumped by all of the time. So I'm gonna show you what that is in a minute so that you don't fall into the same trap. What we mean by a horizontally launched projectile is any object that gets launched in a completely horizontal velocity to start with. So if something is launched off of a cliff, let's say, in this straight horizontal direction with no vertical component to start with, then it's a horizontally launched projectile. What could that be? I mean a boring example, it's just a ball rolling off of a table. If you just roll the ball off of the table, then the velocity the ball has to start off with, if the table's flat and horizontal, the velocity of the ball initially would just be horizontal. So if the initial velocity of the object for a projectile is completely horizontal, then that object is a horizontally launched projectile. A more exciting example. People do crazy stuff. Let's say this person is gonna cliff dive or base jump, and they're gonna be like "whoa, let's do this." We're gonna do this, they're pumped up. They're gonna run but they don't jump off the cliff, they just run straight off of the cliff 'cause they're kind of nervous. Let's say they run off of this cliff with five meters per second of initial velocity, straight off the cliff. And let's say they're completely crazy, let's say this cliff is 30 meters tall. So that's like over 90 feet. That is kind of crazy. So 30 meters tall, they launch, they fly through the air, there's water down here, so they initially went this way, and they start to fall down, and they do something like pschhh, and then they splash in the water, hopefully they don't hit any boats or fish down here. That fish already looks like he got hit. He or she. Alright, fish over here, person splashed into the water. We want to know, here's the question you might get asked: how far did this person go horizontally before striking the water? This is a classic problem, gets asked all the time. And if you were a cliff diver, I mean don't try this at home, but if you were a professional cliff diver you might want to know for this cliff high and this speed how fast do I have to run in order to avoid maybe the rocky shore right here that you might want to avoid. Maybe there's this nasty craggy cliff bottom here that you can't fall on. So how fast would I have to run in order to make it past that? Alright, so conceptually what's happening here, the same thing that happens for any projectile problem, the horizontal direction is happening independently of the vertical direction. And what I mean by that is that the horizontal velocity evolves independent to the vertical velocity. Let me get the velocity this color. So say the vertical velocity, or the vertical direction is pink, horizontal direction is green. This vertical velocity is gonna be changing but this horizontal velocity is just gonna remain the same. These do not influence each other. In other words, this horizontal velocity started at five, the person's always gonna have five meters per second of horizontal velocity. So this horizontal velocity is always gonna be five meters per second. The whole trip, assuming this person really is a freely flying projectile, assuming that there is no jet pack to propel them forward and no air resistance. This person's always gonna have five meters per second of horizontal velocity up onto the point right when they splash in the water, and then at that point there's forces from the water that influence this acceleration in various ways that we're not gonna consider. How about vertically? Vertically this person starts with no initial velocity. So this person just ran horizontally straight off the cliff and then they start to gain velocity. So they're gonna gain vertical velocity downward and maybe more vertical velocity because gravity keeps pulling, and then even more, this might go off the screen but it's gonna be really big. So a lot of vertical velocity, this should keep getting bigger and bigger and bigger because gravity's influencing this vertical direction but not the horizontal direction. So how do we solve this with math? Let's write down what we know. What we know is that horizontally this person started off with an initial velocity. V initial in the x, I could have written i for initial, but I wrote zero for v naught in the x, it still means initial velocity is five meters per second. And we don't know anything else in the x direction. You might think 30 meters is the displacement in the x direction, but that's a vertical distance. This is not telling us anything about this horizontal distance. This horizontal distance or displacement is what we want to know. This horizontal displacement in the x direction, that's what we want to solve for, so we're gonna declare our ignorance, write that here. We don't know how to find it but we want to know that we do want to find so I'm gonna write it there. How about in the y direction, what do we know? We know that the, alright, now we're gonna use this 30. You might want to say that delta y is positive 30 but you would be wrong, and the reason is, this person fell downward 30 meters. Think about it. They started at the top of the cliff, ended at the bottom of the cliff. It means this person is going to end up below where they started, 30 meters below where they started. So this has to be negative 30 meters for the displacement, assuming you're treating downward as negative which is typically the convention shows that downward is negative and leftward is negative. So if you choose downward as negative, this has to be a negative displacement. What else do we know vertically? Well, for a freely flying object we know that the acceleration vertically is always gonna be negative 9.8 meters per second squared, assuming downward is negative. Now, here's the point where people get stumped, and here's the part where people make a mistake. They want to say that the initial velocity in the y direction is five meters per second. I mean people are just dying to stick these five meters per second into here because that's the velocity that you were given. But this was a horizontal velocity. That's why this is called horizontally launched projectile motion, not vertically launched projectile motion. So think about it. The initial velocity in the vertical direction here was zero, there was no initial vertical velocity. This person was not launched vertically up or vertically down, this person was just launched straight horizontally, and so the initial velocity in the vertical direction is just zero. People don't like that. They're like "hold on a minute." They're like, this person is gonna start gaining, alright, this person is gonna start gaining velocity right when they leave the cliff, this starts getting bigger and bigger and bigger in the downward direction. But that's after you leave the cliff. We're talking about right as you leave the cliff. That moment you left the cliff there was only horizontal velocity, which means you started with no initial vertical velocity. So this is the part people get confused by because this is not given to you explicitly in the problem. The problem won't say, "Find the distance for a cliff diver "assuming the initial velocity in the y direction was zero." Now, they're just gonna say, "A cliff diver ran horizontally off of a cliff. "Find this stuff." And you're just gonna have to know that okay, if I run off of a cliff horizontally or something gets shot horizontally, that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. This is good. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity and we're not asked to find the final velocity, we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two. So I get negative 30 meters times two, and then I have to divide both sides by negative 9.8 meters per second squared, equals, notice if you would have forgotten this negative up here for negative 30, you come down here, this would be a positive up top. You'd have a negative on the bottom. You'd have to plug this in, you'd have to try to take the square root of a negative number. Your calculator would have been all like, "I don't know what that means," and you're gonna be like, "Er, am I stuck?" So you'd start coming back here probably and be like, "Let's just make stuff positive and see if that works." It would work because look at these negatives canceled but it's best to just know what you're talking about in the first place. So be careful: plug in your negatives and things will work out alright. So if you solve this you get that the time it took is 2.47 seconds. It's actually a long time. It might seem like you're falling for a long time sometimes when you're like jumping off of a table, jumping off of a trampoline, but it's usually like a fraction of a second. This is actually a long time, two and a half seconds of free fall's a long time. So we could take this, that's how long it took to displace by 30 meters vertically, but that's gonna be how long it took to displace this horizontal direction. We can use the same formula. We can say that well, if delta x equals v initial in the x direction, I'm just using the same formula but in the x direction, plus one half ax t squared. So the same formula as this just in the x direction. Delta x is just dx, we already gave that a name, so let's just call this dx. So I'm gonna scooch this equation over here. Dx is delta x, that equals the initial velocity in the x direction, that's five. Alright, this is really five. In the x direction the initial velocity really was five meters per second. How about the initial time? Oh sorry, the time, there is no initial time. The time here was 2.47 seconds. This was the time interval. The time between when the person jumped, or ran off the cliff, and when the person splashed in the water was 2.4, let me erase this, 2.47 seconds. So 2.47 seconds, and this comes over here. How about this ax? This ax is zero. Remember there's nothing compelling this person to start accelerating in x direction. If they've got no jet pack, there is no air resistance, there is no reason this person is gonna accelerate horizontally, they maintain the same velocity the whole way. So what do we get? If we solve this for dx, we'd get that dx is about 12.4, I believe. Let's see, I calculated this. 12.4-ish meters. Okay, so if these rocks down here extend more than 12 meters, you definitely don't want to do this. I mean if it's even close you probably wouldn't want do this. In fact, just for safety don't try this at home, leave this to professional cliff divers. I'm just saying if you were one and you wanted to calculate how far you'd make it, this is how you would do it. So, long story short, the way you do this problem and the mistakes you would want to avoid are: make sure you're plugging your negative displacement because you fell downward, but the big one is make sure you know that the initial vertical velocity is zero because there is only horizontal velocity to start with. That's not gonna be given explicitly, you're just gonna have to provide that on your own and your own knowledge of physics.

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Horizontally launched projectile (video) | Khan Academy

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Quadratic Equations question 1 solution 2

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Quadratic Equations question 1 solution 2

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Quadratic Equations question 1 solution 1

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Quadratic Equations question 1 solution 1

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https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-numbers-operations/cc-8th-exponent-properties/v/exponent-properties-involving-products

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Welcome to this presentation on logarithm properties. Now this is going to be a very hands-on presentation. If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties. But what I'm going to do is I'm going to show you the properties. And then show you how they can be used. It's going to be little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a-- Oh that's not the right. Let's see. I want to change-- There you go. Let's say I say that a-- Let me start over. a to the b is equal to c. So if we-- a to the b power is equal to c. So another way to write this exact same relationship instead of writing the exponent, is to write it as a logarithm. So we can say that the logarithm base a of c is equal to b. So these are essentially saying the same thing. They just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power you get c. And then you figure out what b is. So they're the exact same relationship, just stated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm-- Let me do a more cheerful color. The logarithm, let's say, of any base-- So let's just call the base-- Let's say b for base. Logarithm base b of a plus logarithm base b of c-- and this only works if we have the same bases. So that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, well I don't know, examples. So this is saying that-- I'll switch to another color. Let's make mauve my-- Mauve-- I don't know. I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of-- I don't know --of 8 plus logarithm base 2 of-- I don't know let's say --32. So, in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Well let's see if that's true. Just trying out this number and this is really isn't a proof. But it'll give you a little bit of an intuition, I think, for what's going on around you. So log-- So this is-- We just used our property. This little property that I presented to you. And let's just see if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see. 2 to the fourth power is 16. 2 to the fifth power is 32. So this right here is 2 to the-- This is 5, right? And 2 to the what power is equal to 256? Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent prop-- I don't know. I don't know the names of things. And that equals 2 to 8, 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 the third times 2 to the fifth, essentially. And on this side, you have them added to each other. And what makes the logarithms interesting is and why-- It's a little confusing at first. And you can watch the proofs if you really want kind of a rigorous-- my proofs aren't rigorous. But if you want kind of a better explanation of how this works. But this should hopefully give you an tuition for why this property holds, right? Because when you multiply two numbers of the same base, right? Two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos. So let's do a-- Let me show you another log property. It's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of-- well I ran out. I'm running out of space --a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of-- I don't know --log base 3 of-- well you know, let's make it interesting --log base 3 of 1/9 minus log base 3 of 81. So this property tells us-- This is the same thing as-- Well I'm ending up with a big number. Log base 3 of 1/9 divided by 81. So that's the same thing as 1/9 times 1/81. I used two large numbers for my example, but we'll move forward. So let's see. 9 times 8 is 720, right? 9 times-- Right. 9 times 8 is 720. So this is 1/729. So this is log base 3 over 1/729. So what-- What does-- 3 to what power is equal to 1/9? Well 3 squared is equal to 9, right? So 3-- So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1/9, right? The negative just inverts it. So this is equal to negative 2, right? And then minus-- 3 to what power is equal 81? 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to-- Well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus sixth power is equal to 1/729. So that's my question. Is 3 to the minus sixth power, is that equal to 7-- 1/729? Well that's the same thing as saying 3 to sixth power is equal to 729, because that's all the negative exponent does is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here. But let's see. 3 to the third power-- This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. You can confirm it with a calculator if you don't believe me. Anyway, that's all the time I have in this video. In the next video, I'll introduce you to the last two logarithm properties. And, if we have time, maybe I'll do examples with the leftover time. I'll see you soon.

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Intro to logarithm properties (1 of 2) (video) | Khan Academy

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Welcome to this presentation on logarithm properties. Now this is going to be a very hands-on presentation. If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties.

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Using the Interactive

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This collection of interactive simulations allow learners of Physics to explore core physics concepts by altering variables and observing the results. This section contains more than 70 simulations and the numbers continue to grow.

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Physics Simulations at The Physics Classroom

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Vectors | Chapter 1, Essence of linear algebra