- Horizontally launched projectile
- What is 2D projectile motion?
- Visualizing vectors in 2 dimensions
- Projectile at an angle
- Launching and landing on different elevations
- Total displacement for projectile
- Total final velocity for projectile
- Correction to total final velocity for projectile
- Projectile on an incline
- 2D projectile motion: Identifying graphs for projectiles
- 2D projectile motion: Vectors and comparing multiple trajectories
- What are velocity components?
- Unit vectors and engineering notation
- Unit vector notation
- Unit vector notation (part 2)
- Projectile motion with ordered set notation
Projectile on an incline
Challenging problem of a projectile on an inclined plane. Created by Sal Khan.
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- I have a big PROBLEM from7:22.Lets say the angel of hill was something much more complicated then 30 degrees,like 73.486 degrees, then we would'nt know the ratio of Sy to Sx...there must be another way...is there ????(15 votes)
- If you don't have a calculator (or sin/cos/tan tables) handy, 73 degrees will be more problematic than 30. If you have a protractor you can draw the triangle with angle 73.something degrees, and measure out SOH CAH TOA using a ruler. You will be limited by the accuracy of your drawing, though!
Some of the most common and useful functions are difficult to compute by hand. Before the calculator, mathematicians (and students) used reference tables of precomputed values.
Logarithms were built on purpose to work that way. Some poor guy spends his whole life building tables, and from then on, the rest of us can just look the numbers up.
If what you meant is "what would I do all by myself?", I think that is a very good question!
For sample exercises, you often use values like 30 degrees because they're easy to work with. When we get the concept, the steps will be the same in a real situation. Some of the intermediate steps may be more tedious, that's all.(65 votes)
- Instead of having a tilted plane that you are shooting off of, couldn't you tilt everything by 30 degrees, making the tilted plane flat, and gravity pulling at a 60 degree angle instead of a 90? All you would have to do is use trig to find that instead of gravity having a force of -9.8, it had a force of -4.9 square root of 3.
Would this work?(28 votes)
- Yes, you are right. The problem can be solved that way too. However, it gets slightly more complicated. This is the simplest way.(5 votes)
- If you factor in air resistance, how would that change the problem? And how would you solve it?(19 votes)
- If you include air resistance, you would have an x-direction acceleration. There would be an equation for it, but not a kinematic since air resistance varies with speed. To describe it simply, the object would constantly be slowing down, but at a constantly decreasing rate.(5 votes)
- Can we assume our frame of reference with respect to the inclined plane???
I mean.. our inclined plane will be the x axis.. in this case.. the condition for the tootal distance travelled will be y co-ordinate will be 0...????
Then we can simply use projectile motion formulas by using theta as angle with ground -angle of inclined plane with ground???(8 votes)
- You can, but... The acceleration due to gravity will now have two components and not be straight down.(12 votes)
- Why don't we take the incline as the horizontal axis? That way the horizontal displacement alone should be our answer, shouldn't it?(9 votes)
- It's too bad there aren't videos on axis rotation, I often think about how to change a problem in this way but haven't really looked at how it works. Anyway, having gravity not be vertical seems like it would complicate things counter-productively!(9 votes)
I solve for time with tan (30) = (7.07t-4.9t²)/(7.07t) => t=0.83
But when I want to figure out the horizontal component Sx then:
7.07*t = Sx
But: 7.07*0.83 = 5.86
and Sal has 4.31
I don't understand why(5 votes)
- Ok, I found my mistake there: tan (30) = (7.07t-4.9t²)/(7.07t) => t=0.609(16 votes)
- Can anyone explain how we could do it if we take the same hill but we fire the porjectile from its top? With the same angle as before in the example above.(7 votes)
- At around13:34why does the first S sub y end up becoming 1? Thank you(6 votes)
- The earlier equation was:
S sub y=sqrt(3)S sub y - 4.9 (3*S sub y/50)
Now subtracting S sub y on both sides:
0 = sqrt( 3)S sub y - S sub y - 4.9(3*S sub y/50)
In the expresion - (sqrt(3)S sub y -S sub y), we can take S sub y common...
so it becomes,
0 = S sub y(sqrt(3) -1) - 4.9*(3*S sub y / 50)
Thank You(11 votes)
- At5:45we can see that the Vertical Displacement and the Horizontal Displacement are not the same functions. The Sx is missing the -4.9Δt². Why is that?(3 votes)
- I just thought that it could be because the gravity doesn't affect the horizontal displacement.. Is that the reason?(1 vote)
- Why again is there a "minus one" at13:34?(4 votes)
- Sal had to set his equation equal to zero...so when he subtracted the Sy from both sides.....his left side gave him a zero....and on this right side it gave him sqrt(3)*Sy minus 1*Sy....to make it pretty and exact Sal wrote it like this.....(sqrt(3)-1)Sy....(factored the Sy so we could look at the numbers......Basically, all he did was collect like terms and put them on the left side of the equation and set it equal to zero.(2 votes)
What I want to do in this video is tackle a problem that would be considered pretty difficult for most first-year physics students. And you, frankly, probably wouldn't be expected to solve a problem like this in most first-year physics class. Or if you're in an advanced or honors class you might be expected or it might be a bonus problem. But it's an interesting type of problem. Because what we're going to do is, we're going to launch a projectile on an incline. So maybe we're on the side of the hill. So it's a hill. Let me do it in green. So let's say we're on the side of the hill like this. And let's say that we know the inclination of the hill. The hill's inclination is 30 degrees off the horizontal. So this is the horizontal right over here. So that is the inclination of the hill. And then we're going to launch a projectile at 10 meters per second. We're going to launch it at 10 meters per second. And the angle with the hill is 15 degrees. So at a 15 degree angle with the hill. And the reason why this is more difficult than the traditional projectile motion problems is, well, we could think about it. The projectile is going to be launched and it is going to eventually land at some point on the hill. But we can't do the simple figure out how long it's in the air using its vertical velocity because we don't know what the vertical displacement for this thing is going to be, unless we know how far down the hill it lands. Because the further down the hill it lands, the higher the vertical displacement. So we have to think about both the horizontal and the vertical displacement at the same time. And as we walk through this, you'll see how that can be done. So I guess the first thing that we really would always want to do whenever you want to try to solve this type of problem is break up our velocity into both the horizontal and vertical components. So the vertical component of our velocity is going to be the magnitude of our total velocity, 10 meters per second, times-- and be very careful here-- not the sine of 15 degrees, but the sine of the angle with the horizontal. So times the sine of 45 degrees. And I go into it in a lot more detail in previous videos. For the sake of time, I won't go into it. But it really just comes from sohcahtoa. If we were to draw the vertical component, it would look like this. This is the angle. The sine of 45 degrees is equal to the opposite over the hypotenuse or the hypotenuse times the sine of 45 degrees is equal to the vertical component. That's where it's coming from. Let me get rid of some of the stuff I just drew just so it makes it a little bit cleaner. And the horizontal component of our velocity is going to be, by the same logic, 10 cosine of 45 degrees. Now let's think about what the horizontal displacement is going to be. And I'm just going to go straight to the formula that we've derived in the last few videos. The horizontal displacement is going to be our initial-- sorry. Let's do the vertical displacement. The vertical displacement-- I could have done the horizontal displacement first-- but the vertical displacement is going to be our initial vertical velocity, which we know as 10 sine of 45 degrees. And by the way, we could just solve that right now. What is the sine of 45 degrees? Sine of 45 degrees is square root of 2/2. Cosine of 45 degrees is also square root of 2/2. So both of these values, 10 times square root of 2/2 is 5 square roots of 2. So this whole thing right over here is 5 square roots of 2 meters per second. That's our vertical velocity. And our horizontal velocity is also 5 square roots of 2 meters per second. So that simplifies things a little bit. But anyway, we were talking about our vertical displacement. Our vertical displacement is going to be our initial vertical velocity, 5 square roots of 2, times our change in time plus the acceleration. Well, we know what the acceleration is. It's negative 9.8 meters per second squared. So let me write minus 9.8. I'm not going to write the units here so that we say space. Times our change in time squared. All of that over 2. We derive this in several videos, especially the last few where we do these two dimensional projectile motions. So this gives us our displacement in the y direction. And I can simplify this a little bit. Our displacement in the vertical direction is equal to 5 square roots of 2 times delta t times-- let me do it in the same-- times delta t change in time minus 4.9 times our change in time squared. So we know we have this constraint right over here. So this gives us our vertical displacement as a function of time. Let's think about our horizontal displacement as a function of time. Our horizontal displacement is going to be equal to our horizontal velocity, which is 5 square roots of 2 times our change in time. Now what can we do next? Well, we have to have some relationship between our horizontal displacement and our vertical displacement. And that relationship is going to be given to us by this incline. So wherever we land-- let's say this is where we eventually do end up landing. Well, let's think about our horizontal and our vertical displacements and what their relationship would have to be. So if this is where we land, then this would be our-- let me do it in the same colors-- that right there would be my vertical displacement. I would move that far up. And then our horizontal displacement will be this right over here, will be that length right over there. So that is our horizontal displacement. So what is the relationship between our vertical displacement and our horizontal displacement? And we know that this angle right over here is 30 degrees. So we can use some basic trigonometry. We have a right triangle. We know the opposite side from the angle. We know the adjacent side. And the trig function that uses the opposite and the adjacent is the tangent function. So we get the tangent of 30 degrees is going to be equal to the magnitude of our vertical displacement over the magnitude of our horizontal displacement. And the tangent of 30 degrees, that's the same thing as the sine of 30 over-- let me just do it over here. So the tangent of 30 degrees is the same thing as the sine of 30 degrees over the cosine of 30 degrees. Let me do this a little bit neater. And the sine of 30 degrees is 1/2. This is equal to 1/2. And the cosine of 30 degrees is the square root of 3/2. So this is equal to 1/2 times 2 over the square root of 3, which is equal to 1 over the square root of 3. So we get the magnitude of our vertical component over the magnitude of our horizontal component-- is our horizontal component right over here-- is equal to 1 over the square root of 3. What's useful about this is it gives us a relationship between our horizontal and vertical component, or between our vertical and our horizontal components. And we can use this constraint right over here to then solve for one of these two. And let me show you how we'll do it. So let's just explicitly write this. Well, let's do it this way. Let us cross multiply here, which is really the same thing as multiplying both sides by the square root of 3 and the magnitude of our horizontal component. We get the square root of 3 times the magnitude. And both of these are going to be positive. Well, let me just write it this way. Times the magnitude of our vertical component is going to be equal to the magnitude of our horizontal component. So right, just like that. So we now have a relationship between the length of these two vectors. And we can use this relationship to substitute back into the constraints that we already have. So the second constraint right over here, let's use this information. The second constraint says that our horizontal component of our displacement is equal to 5 square roots of 2 times our change in time. Or another way of thinking about it, if we divide both sides by 5 square roots of 2, we get our change in time is equal to the horizontal component of our displacement divided by 5 square roots of 2. But we also know that the horizontal component of our displacement is the square root of 3 times the vertical component of our displacement. Here I explicitly wrote the magnitude notation. When we start dealing with either just the vertical or the horizontal component, I can just write it like this, because it's either going to be a positive or a negative value. And that specifies both the magnitude and the direction. So what I'm going to do right over here, and obviously the way I've drawn it right over here, both of these are going to be positive values. It's upwards displacement in the vertical direction, so that's positive by our convention. And we're moving to the right. So that's positive, also, by our convention. So I can rewrite this over here as being equal to square root of 3 times our vertical displacement. And all of that's over 5 square roots of 2. Now, the whole reason why I did this is, this expression right here contains this information, contains the ratio between our vertical displacement and our horizontal displacement. And it also contains the information of, how does the horizontal displacement, how does that change as a function of time? So our time needs to be equal to this. So this is our time as a function of our vertical displacement now, not time as a function of our horizontal displacement. And what we can do is, we can use this constraint with our original vertical displacement as a function of time to then solve for our vertical displacements. So let's do that. Let's substitute this business right here for delta t in this top equation right over here. So if we do that-- and I'll write it big-- we get our vertical displacement, right over there, is equal to 5 square roots of 2 times delta t. So it's 5 square roots of 2. Delta t is all of this business over here. So 5 square roots of 2 times delta t. Delta t is square root of 3 times our vertical component. All of that, really the magnitude of our vertical component, over 5 square roots of 2. So that's that right there. Or actually, we could look at this one right over here. We could use this constraint. This is just simplified. So then we have minus 4.9 times delta t squared. So delta t squared is this quantity squared. I'll just write it out. I don't want to skip too many steps. So delta t once again is the square root of 3 times the vertical component. All of that over 5 square roots of 2 squared. And now what does this give us? So now we literally have a quadratic equation with only one variable. So we can solve for this. But let me rewrite it. Let me simplify it. So we have our vertical component is equal to-- now we have a 5 square roots of 2 in the numerator and one in the denominator and they cancel out. So we get the square root of 3 times our vertical component, the magnitude of our vertical component. It's actually also specifying, well, the magnitude, we can say, for now, although I'm not using that notation. So then we have minus 4.9 times this quantity squared. So that's going to be, the square root of 3 squared is 3 times the vertical component squared. And then over 5 squared, which is 25, times 2. That's the square root of 2 squares. So 25 times 2 is 50. And so we get, if we were to simplify this a little bit more. Let's subtract this from both sides. We get-- I'll do it all in one color-- 0 is equal to the square root of 3 minus 1 times our vertical component. Because if we subtract this from both sides, that's square root of 3 times our vertical component minus 1 times our vertical component. So that's the square root of 3 minus 1 times our vertical component. And then we have all of this business. Minus 4.9 times 3 over 50 times our vertical component squared. And lucky for us, we can just factor out one of these s sub y's over here, one of these vectors. And so we can get-- and I'll just do that in place so that I don't-- well, let me just-- I don't want to skip many steps. So this is equal to the square root of 3 minus 1, minus 4.9. Let me do it in that color. So let me do it like this. It's equal to the square root of 3 minus 1 minus 4.9 times 3 over 50 times one of these, times our vertical component. And then we factored one of those vertical components out. So we factored one of them out. So the vertical component of our displacement could either be-- so we have the product of two things that equal 0. So our vertical displacement could be 0, which is true, because at some point in the path we literally had 0 vertical displacement. That was literally where we started. But that's not the answer we're looking for. We're looking for this vertical displacement. So either this is going to be 0. But that's just kind of the obvious answer. Or all of this business is going to be equal to 0. But this is pretty easy to solve for 0 over here. So we get square root of 3 minus 1 minus 4.9. Let me just calculate all of these things just so that I don't have to keep writing them. So we get the square root of 3 minus 1 is equal to 0.73205. So I'll just write 0.732 here. So this is equal to 0.732. That's that part right over there. And then 4.94. I know this problem is getting long. Just pause it and take a break if you're getting tired. 4.9 times 3 divided by 50 is equal to 0.294. So minus 0.294. I could put a 0 out front just to make it clear where we are. Times this, times our vertical component. This could also be equal to 0. Either this is 0 or that is 0. When this is 0, it gives us the obvious answer. We're more interested in this. To solve for this, we can add this to both sides. And we get 0.732 is equal to negative 0.294 times the vertical component. We are in the home stretch. We divide both sides by this, by negative 0.29. Oh sorry, this will now be positive. I almost made a careless mistake after 16 minutes of video. So now we divide both sides by 0.294 and we get our vertical displacement. So this, I think, a drum roll might be in order. So we have 0.732 and all of this business, but I'll just round it there, divided by 0.294 gives us a vertical displacement of, if we round, 2.50 meters. Or 2.49 meters, I should say. So this is equal to 2.49 meters. This is exciting. This is equal to 2.49 meters. And now we can figure out the horizontal displacement pretty easily, because we know that the horizontal displacement is square root of 3 times the vertical displacement. So let's figure that out. That's the vertical displacement. Let's multiply that times the square root of 3. And we get 4.31 meters. So we get the horizontal displacement. The horizontal displacement is equal to 4.31 meters. So this is equal to 4.31 meters. So we actually now know the total displacement in the vertical direction and in the horizontal direction. And I'll leave it up to you. If you wanted to figure out exactly how far along the hill we traveled, you can just use both of these values in the Pythagorean theorem to essentially figure out the hypotenuse of this right triangle.