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Launching and landing on different elevations

More complicated example involving launching and landing at different elevations. Created by Sal Khan.

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  • piceratops ultimate style avatar for user Jayati
    I understand that time can't be negative for the purposes of this video, but can it be negative in any other situation? Please specify an example.
    (54 votes)
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  • mr pink red style avatar for user Dale Chapman
    My only issue with this video is that you didn't specify the length of the 'well' in the middle. How do I know that the projectile won't land there instead of the second plateau?
    (43 votes)
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    • blobby green style avatar for user David Garcia
      In other words, we are told that the projectile lands on the second plateau. That is a fact. What we could ultimately answer is the magnitude of the horizontal displacement (i.e. the distance from the cannon to the place where the projectile lands). The length of the 'well' will remain forever a mystery!
      (27 votes)
  • blobby green style avatar for user Meena Ahmed
    @, why didn't Sal use the formula: s = Vi(x) + a(deltaT)
    is it because the acceleration -9.8m/s^2 only applies in the vertical direction?

    I'm just confused because I thought V(x) of 90cos53 was only for Vi(x) and not Vavg(x)

    Thanks for answering!
    (8 votes)
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    • blobby green style avatar for user Johnny Cantrell
      Sal, worked this problem in such a way to avoid using several other algebraic equations. He knows that Juniors and Seniors in high school and college know the quadratic equation. So he solved for the time by this means. There is another way (using basic algebra....takes forever)...For Y(direction)....You will need to use....Vf=Vi+at (Vf=0 Vi=90sin53 a=-9.8) t=7.33(this is just the time up) but total air time would be (7.3344*2= 14.67s)..to land on its original horizontal axis...and a=-9.8 because the object is going up and gravity is slowing its upward velocity...to where Vf=0 at 7.3344s)....Now as the objects falls, since the y-direction changes...its as if Vi=0 and Vf=90sin53...as it returns to its horizontal reference...BUT once it falls below it....we have a new calculation....to consider....Our Vi=90sin53 because it has an initial velocity but we now do not know its new final velocity...We have to find it.... Vf^2=Vi^2+2ad (Vi=90sin53 and a=9.8 and d=16) so Vf=74.03m/s..Now its is time to find how long it took to fall this 16m.......then Vf=Vi+at (Vf=74.03 Vi=71.88 a=9.8)and a=+9.8 because gravity is helping its Velocity increase....thus....t=0.22s (time it takes to fall the 16m)....final time in the air will be 14.67s+0.22= 14.89s....then to get the X-distance just use D=V*t...(D=90cos53*14.89= 806.5m....Sorry for the long winded explanation.
      (22 votes)
  • blobby green style avatar for user rougiekat
    Very stupid question:
    Is the horizontal component of a velocity vector, how far the object was displaced or the total time it was in the air?
    (4 votes)
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  • blobby green style avatar for user beepcool979
    how come all the equations used are only applied to the vertical velocity of the cannonball?
    (5 votes)
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  • blobby green style avatar for user yagamiri
    hi, so does this mean that time up = time down also applies here when the elevation is different? sorry and thanks!
    (2 votes)
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  • piceratops ultimate style avatar for user Anay
    Sal has solved this using Quadratic formula, but I don't understand it is there any other way solve it?
    Like by factoring or something?
    (4 votes)
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  • duskpin ultimate style avatar for user areej
    hello . i want to ask why did sal put 90sin53 as a initial velocity instead of 90m/s , can someone explain it more?
    (4 votes)
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  • starky sapling style avatar for user King Keilogy
    how come 2*negativeA gives another negative value
    (3 votes)
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  • leaf green style avatar for user maungpyaichan
    Hi, I don't understand why the horizontal velocity is constant. Is it always this case? Thank you.
    (3 votes)
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    • leafers sapling style avatar for user Agnivo Paul
      Horizontal velocity is always constant, when the air resistance is negligible. With an acceleration, the component of velocity cannot change. Vertical velocity is affected by gravitational pull (9.8m/s). But the horizontal velocity, faces zero negative or positive acceleration in the mid air, until you are talking of a self-accelerating body like flight. So generally, the horizontal velocity is taken constant.
      (3 votes)

Video transcript

Let's do a slightly more complicated two-dimensional projectile motion problem now. So in this situation, I am going to launch the projectile off of a platform. And then it is going to land on another platform. And I'm going to fire the projectile at an angle. Let me draw this a little bit better. So I'm going to fire the projectile at an angle of-- let me use a not so clean number. Let's say it is a 53-degree angle. And it's coming out of the cannon. And let me make it clear. So it's coming out-- let me do it this way, just to make it 100% clear. So this angle right over here is 53 degrees. And we are going to have it come out of the muzzle of the cannon with a velocity of 90 meters per second. And just to give ourselves a sense of the heights, or how high it's being launched from-- so from the muzzle of the cannon down to here. So this height right over here, let's say that that is 25 meters. And let's say that this height right over here is 9 meters. And so we're essentially launching this from a height of 25 meters. I know in the last video, even though I drew the cannon like this, we assumed that it was being launched from an altitude of 0 and then landing back at an altitude of 0. Here we're assuming we're launching it from an altitude of 25 meters, because that's when it's leaving the muzzle. And it's going to start decelerating at least in the vertical direction as soon as it leaves the muzzle. And then we're assuming it's not going to land back at the same altitude. It's going to land at a different altitude. So how do we think about this problem? So the first thing you'll always want to do is divide your velocity vector into its horizontal and vertical components. You use the vertical component to figure out how long it's going to stay in the air. And then you use the horizontal component to figure out, given how long it's in the air, how far did it travel. And once again, we're going to assume that air resistance is negligible. So just based on what we did in the last video-- and I'll go through all of the steps in this one, as well. So if we draw our vector, the length here is going to be 90. This is our velocity vector. The angle over here between the x-axis and our vector is 53 degrees. And let me draw the horizontal component. The horizontal component would look like this. And the vertical component-- I'll do it in orange. The vertical component would look like-- that's not orange. The vertical component will look like this. And so the vertical component of the vector-- what would be the length of this side right over here? Well, this is the opposite side. We know from basic trigonometry sine of an angle is opposite over the hypotenuse. So we know that the sine of 53 degrees is equal to this opposite side, is equal to the vertical velocity. And I could write it's the magnitude of the vertical velocity. I write that subscript y, because we're in the y direction. That's the vertical direction, over the length of the hypotenuse, over the magnitude of our original vector. Or we can get that this side right over here. If we multiply both sides by 90, we get that the magnitude of that side is going to be equal to 90 times the sine of 53 degrees. Now, if we want to do the horizontal component, the horizontal side is adjacent to this. Cosine, soh cah toa. Cosine is adjacent over hypotenuse. So the horizontal component of our velocity, I'll say in the x direction, over the hypotenuse, over 90, is equal to the cosine of 53 degrees. Cosine is adjacent over hypotenuse. Adjacent, that's this length, over 90. Multiply both sides by 90. You get that the horizontal component is equal to 90 times cosine of 53 degrees. Now, how do we figure out how long this thing stays in the air? Well, we'll use the vertical component for that. And especially since we're dealing with different levels, we can't use that more basic reasoning, that hey, whatever velocity we start off at, it's going to be the same velocity but in the opposite direction. Or the same magnitude and velocity but the opposite direction. Because we're not going to the same elevation. But what we could do is we can use the formula that we derived in the previous video. That the displacement-- let me just copy and paste this a little bit lower so we can use it. So I'll stick it right over here. So we could use this. We know that the displacement is equal to the initial velocity-- and we're dealing with the vertical direction right here-- times the change in time plus the acceleration, times the change in time squared, divided by 2. So how do we use this to figure out how long were in the air? So what is the displacement? If we're starting at 25 meters high, and we're going to 9 meters high. So over the course, while this thing is traveling, it will be displaced downwards 16 meters. Or another way to think about it is our displacement in the vertical direction is going to be equal to negative 16 meters. Let me write that a little bit bigger. Negative 16 meters, right? Because 25 minus 9 is 16. And so we can put that into the formula that we derived in the previous video. We get negative 16-- I won't write the units here just so that we don't take up too much real estate, so that at least it looks simple-- is equal to the initial velocity. We're dealing with just the vertical dimension here. So we're just dealing with the vertical. And remember, it's negative because our displacement is going to be downwards. We're losing altitude. So our vertical velocity-- we already figured that out. It is 90 times the sine of 53 degrees. Actually, let me do it in that same color. The first time we do a new type of problem, it's good to know-- so 90 times the sine of 53 degrees, times our change in time, is equal to the acceleration of-- due to the force of gravity for an object in free fall is going to be negative 9.8 meters per second squared. But we're dividing that by 2. So we're going to have minus 4.9 meters per second squared times delta t squared. Times our change in time squared. So how do we solve something like this? You can't just factor out a t and solve it. But you might recognize this is a quadratic equation right over here. And the way you solve quadratic equations is you get everything onto one side of this equation. And then you either factor it out. But more likely in this situation, we will use the quadratic formula, which we've proved in other videos and hopefully given you the intuition for it-- to actually solve for the times where your elevation, where your displacement in the vertical direction is negative 16 meters. And you'll get two solutions here. And one of the solutions will be a negative change in time. So it'll be like at sometime in the past, you were also at negative 16 meters. That's nonsensical for this problem. So we'll want to take the positive value here. So let's put all of this on one side of the equation. So let's add 16 to both sides. On the left-hand side, you just get a 0. 0 is equal to-- and I'll write it in kind of the traditional way that we're used to seeing it. I'll write the highest degree term first. So negative 4.9 times delta t squared. And then we have plus 90 sine of 53 degrees times delta t, and then plus 16. I'm going to do that in yellow. All of this is equal to 0. And this, once again, is just a quadratic equation. We can find its roots. And the roots will be in terms of delta t. We can solve for delta t using the quadratic formula. So we get delta t-- and if this is very unfamiliar to you, review the videos on Khan Academy in the algebra playlist on the quadratic formula. And if you don't know where it came from, we also prove it for you. So it's equal to negative B. B is this term right here, the coefficient on the delta t. So it's going to be negative 90 sine of 53 degrees. I'll write the quadratic formula up here for those of you who don't fully remember it. So if I'm trying to solve Ax squared plus Bx plus C is equal to 0, the roots over here are going to be negative B plus or minus the square root of B squared minus 4AC, all of that over 2 times A. These are going to be the x values that satisfy this equation up here. So that's all I'm doing over here. This is the B value. Negative B plus or minus-- and it's going to turn out that we only care about the plus one, because that's going to give us the positive value. But I'll just write it out here. Plus or minus the square root of B squared. So it's this quantity squared. So it is 90 sine of 53 degrees squared, minus 4-- we're going to run [? out that ?] little space. So minus 4 times A, which is negative 4.9. Well, let me just write negative 4.9 times C. C over here is 16, times 16. Let me put this radical all the way over here. All of that over 2A. So A is negative 4.9. 2 times A is negative 9.8. So now we can get the calculator out to figure out our change in time. And I'm just going to focus on the positive version of it. I'll leave it up to you to find the negative version and see that it'll give you a negative value for change in time. And that's nonsensical, so we only care about the positive change in time where we get to a displacement of negative 16 meters. Let's get the calculator out. So we get-- want to do this carefully. We have negative 90 sine of 53 degrees plus. I'm doing the plus version here because that'll give us the positive value-- plus the square root of. And I'll do this in parentheses. 90 sine of 53 degrees squared. That's that part right there. These two negatives cancel out. So I could say this is plus 4 times positive 4.9. So plus 4 times 4.9 times 16. And then that closes off our entire radical. And so this will give me the numerator up here. That gives me the numerator. And then I want to divide that by-- did I do the negative 90? Oh, and I just realized that I made a mistake. I said that the positive version would give you the positive time. But now we realize that's wrong. Because when I took the positive version, when I put a plus up here, I get a positive 2.14 for the numerator. But then we divide it by negative 9.8. We're going to get a negative value. So that's not going to be the time that we care about. We care about the time where this is a negative value. So let me re-enter that. So let me do the negative value. So let me move back a little bit. And then let me replace this with a minus sign. So I'm going to look at the negative value, because I want the positive time. And so now my numerator here is a negative value. And so this is actually what we care about. We care about the numerator's a negative value. You divide by negative 9.8, and you get 14.8-- I'll just round-- 14.89 seconds. So delta t, the positive version, is equal to 14.89 seconds. And so my initial comment about using the positive version was wrong because we have this denominator that's negative. So you want the numerator to be negative. And only when the numerator is negative will the whole expression be positive. And so we got this positive time of 14.89 seconds. I'm going to leave you there. And the next part of the video-- actually, I might as well just solve it instead of making a new video. Although this is running long. So the amount of time that we're in the air is 14.89 seconds. So if I were to ask you the horizontal displacement, it's going to be the amount of time we're in the air times your constant horizontal velocity. And we already figured out our constant horizontal velocity. So if you want to figure out how far along the x-axis we get displaced, we just take this time times-- that just means our previous answer-- times this value right here, times 90 cosine of 53 degrees. And that gives 806 meters. So this displacement right over here is 806 meters.