In this video, I go over projectile motion along an inclined plane and show that the launch angle that gives the maximum range upwards or downwards is half the angle between the plane and the verti...
Transcript
The solution to problem two, and this one is involving a projectile fired down an inclined plane. So part A states a projectiles fired from the origin down an inclined plane that makes an angle theta with the horizontal, and the angle of elevation of the gun initial speed of the projectile are alpha and v0, respectively, and rest find the position vector of the projectile and the parametric equations of the path of the projectile's functions of time T and ignore air resistance. So here's the projectile. So again, there's the alpha, there's the launch angle, there's the launch speed, and then there's the theta, which is the inclination angle with the horizontal and this slope right there. and basically we want to find parametric equations and, yeah, the vector equation of this, or the position vector of this path right here as it goes down here. That's fascinating stuff. So let's take a look at Part B, read it out, and then I'll go over Part A, solution, and so on. So part B states show that the angle of elevation alpha that will maximize the downhill range is the angle halfway between the plane and the vertical. Yes, yes, that's pretty interesting. So if you look at it, so of the plane, so this is the plane, Halfway would be, and this is a vertical, so halfway is gonna be somewhere like that. So if you want to maximize the downhill range, so how far it goes downhill, this is gonna be the farthest one. So halfway, not just 45, you need to go halfway of this actual hill. That's pretty interesting. So it's similar to when it's flat, you just shoot halfway, which is 45 degrees. That's pretty interesting. And now Part C states, suppose the projectile is fired up. So instead of down, we're going to firing up, an inclined plane whose angle of inclination is data. So we're going to be firing it up. That's theta, well, we're going to go with theta, I guess. So basically, we're launching it this way. and now it's saying show that in order to maximize the uphill range, projectile should be fired in a direction halfway between the plane and the vertical. So same thing, but now, so it's going to be, this is the vertical, so we need to fire, this is the plane, so it's going to be fire halfway between these two. So that's going to be it to maximize how far up it goes there. So that's pretty unique setup there. Alright, and then Part D states, in a paper presented in 1686, Edmund Haley summarized the laws of gravity in projectile and applied them to gunnery. One problem he posed involved firing a projectile to hit a targeted distance R, up an inclined plane, show that the angle at which projectile should be fired to hit the target, but used the least amount of energy is the same as the angle in Part C. and it says, use the fact that the energy needed to fire the projectiles proportional to the square of the initial speed. So, minimizing energy is equivalent to minimizing the initial speed. So basically, even just to minimizing energy, do fire a target here, you would still need to just fire the same as Part C, which is halfway between the vertical. So again, it's like a lot of physics going on evolve. So you still need to go halfway between these two, even just to minimize energy. So instead of firing it up like this, it will launch it like that. So the farthest distance, will also minimize the energy, so basically you're going to have the lowest amount of speed necessary to hit the target. All right, fascinating stuff. Alright, so now let's take a look at the solution to Part A. So part A, again, I'll read it again, so projectiles fire from the origin down in inclined planes, we're firing down like this. Yeah, that makes an angle theta with the horizontal, so there's theta. and the angle of elevation of the gun and initial speed of the projectile are alpha and v0, respectively, and find a position vector of the projectile and the parametric equations of the path of projectile's functions of time T. So again, we're just firing, this is again part A, we're just firing it like that, there's alpha. there's theta and so on, there's incline plane, so we want to know the path there. So, so solution, so as in problem one, and then I'll go all the way back up to part A, where we just had the position vectors of a, just a projectile from my earlier video. So I'll go all the way back up to problem 1A, and it says 1B, so yeah, this problem 1 was quite extensive. So here's a projectile, this is a path of projectile. Alright, so that's the path of the projectile. I'm going to copy this. And then, yeah, we're going to work with that and basically simplify it and find our specific parametric equations for it, as opposed to just using a generic one, like that. Okay, so we have them. I'm going to plug this inside here. All right, so there's our equations. All right, so that is, yeah, so R, the position vector, vector equation, R of T, equals 2, equals v0, Kosa, alpha, T, and then I vector, plus v0, sine alpha T minus 1 1, gt squared, and then the J standard-based vector. So, in other words, this is the X, that's the Y, so we get, so, X component's going to be v.0, cosine, alpha, T, and then y is equal to v0 sine alpha t minus one half gt squared Alright, so this is the parametric equations, just the generic ones of the projectile, but the difference here is that the projectile travels until it reaches a point where, basically, yeah, we have it, we're going down to incline, so I just copy and paste it that, instead of drawing my own. So it goes there, so here's the x and y axis, there's the zero. So then at the point here, let's say it hits this point there. So at this point, this is going to be our y, this is going to be our x. So, yeah, this is going to be our X, like that. So, what that means is, well, if we just take the tangent right here, first of all, yeah, we have X is greater than 0. So we have X is greater than, yeah, grade in zero, and we have Y. Yeah, yeah, so basically the projectile travels until it reaches a point where X is graded in zero and Y is going to be over here, so we use the tangent setup there. So we have tangent theta. is equal to opposite over adjacent, so that's going to be our y over x. So basically y over x, move this over to the other side, but it's going to be negative now. So it's actually going to be negative, yeah, negative, yeah, I'll put a negative y, yeah, like that, so negative y, like that, just because we were going down, downhill there. So that y is y is actually going to be negative, yeah, I'll put a negative y like that, because it's going down here. So that's a negative y, And now what we'll do here is, yeah, we'll do is move this x over the other side. So basically, what we're going to get, move it over there, so on the top. So what we get is y, and also move the negative as well. So y is equal to negative 10 theta times by x. And also I'm going to write, well, here's also, we also have is that this theta, where positive, we'll just go positive, make the whole thing positive. It's going to be Theta is between 0 and pi over 2. So between 0 and 90 degrees. So we're not going to make it go behind us. So it's going to be like that. This is going to be 90 degrees max or pi over 2. Radians. Yeah, so just leave it like on. All right. And now from the parametric equations we obtain, it's again similar to problem 1. So we take this right here, x there. We're going to move this over to the other side. So move this over to the other side, we get T, and then we take the T, and then we throw that inside all that. So what we get is, and I'll do that all right away, because we did this before, we're going to get T is equal to x over v0, and so v0 cosine alpha, and then throw that inside the y, we get y is equal to, uh, right here v, uh, sign alpha, and then t, v. not sine alpha and then t is going to be well just going to be x over v not cosine alpha so that's our t minus one half gt squared so g two squares is going to be well x over v not cosine alpha and then squared like that All right, so we have that, and likewise, we're going to do some cancellations, right here, cancel v knotss, and then we're going to have the sine alpha, alpha over cosine alpha, that's just tan alpha. So we get, so it's going to be y is equal to tan alpha, x, and then minus now we have this on the top. We're going to have a GX squared over 2v0 squared. cosine squared alpha. All right, so let's, now we have this, so thus the projectile hits the incline plane at a point where, so that's parametric, or just that's the y, equation of it, the parametric equations, but in terms of y, there, but we also know that it hits this point right here, y equals negative tan x, so this equals two, yeah, that's just a negative there, and here, just for clarity, I'm just put this up, Positive Y there, yeah, but this still holds it. We're going to go on the negative side, just so that we have Y here is equal to negative tan. Put it like that, Theta X, like that. So that's what we have, just because the opposite, it was downhill, because we usually do the opposite of a descent from the positive side there, and so on. But anyways, keep it like that, just so we have this Y equals negative tan, Theta, X. So that's where it hits. So, in other words, This equation has to equate to it, so thus the projector that's the incline plane at a point where we have y is equal to 10 alpha x minus g x squared 2v not cosine squared alpha equals to negative and now this is 10 theta notice this one is alpha now we have a theta and then X like that. So what we can do now is, what we can do is if we move this over to the other side, move this over to the other side, just so we have positives everywhere, and then put this one first. So we have GX squared over 2v0 cosine squared alpha. Now this equals 2, and we're going to factor out the x. Yeah, so we're going to factor that out and put on the right side, so we're going to have a tan alpha. plus 10 theta, and now this is going to be, well, x. And since, yes, since x is greater than 0, so that's why if we can cancel this out, this out, because number, this one's always positive, x squared is always positive, so when we have to be careful, because we can't necessarily cancel, we might have a negative positive and so on. But since this is positive, we could just cancel that out. Yeah, so since it's positive, then, yes, then what we have is, well, we can have a GX like that, just cancel out the, yeah, so basically x is positive, it doesn't matter, we don't need to worry about any negative side. So 2v squared, v0 squared, cosine squared alpha equals to 10 alpha plus 10 theta. Alright, now the next thing we could do is, well, we'll solve for X. We already solve for Y, so now what we'll do is, well, Y in terms of X. But now what we could do is move this all the way to the other side. So we're going to have an equation for X, so thus, x is equal to, yeah, to basically move this, yeah, this one's going to be on the top, the G is going to be at the bottom, so we're going to have a 2 v0 squared cosine. S squared alpha over G and then times it by 10 alpha, yeah, 10 alpha plus 10 theta, like that. All right, so we have this, but again, remember, we were asked to make parametric equations for part A in terms of T, so it's functions of T, and we already have T over here, so X equals to v0, cosine alpha T, so we already solved this over here. So what we'll do is, so, and, yeah, so, and T is equal to, let's put that again, what is that, x over v .0, cosine alpha. X equals to v. co, yeah, I mean, x over, t is equal to x over v0 cosine alpha. So now we just plug in our x inside, yeah, plug in this whole setup inside, and then we're going to get, yeah, we're going to get over here, so notice here, we're going to have a v0 squared, so that's going to cancel with this, so we're, yeah, that cancel it, and this, this squared cancels with this, so we'll have a G at the bottom, and then, yes, this will equal to two, v0, and then cosine alpha, so the square is gone, now we just divided by G. Alright, so let's put this alpha like that. So that's what t is equal to, and this is going to be 10 alpha plus 10 theta, like that. All right, so what this means is, well, the parametric equations that we already define, so we have x is equal to this setup right here. Yeah, over here, so we have x over here, or in terms of just the generic ones here, so we have this x equals v0, cosine alpha t, and this one's why is this, but now the interval for t is going to be, so that's the equation for t are in the interval from zero, and then all the way to its maximum, this is when it hits the ground at the inclined plane, it's going to be to v0. cosine alpha over G, and then 10 alpha plus 10 Theta. So that is the T value from beginning to the end. So initially T equals 0, then T equals to all that when it hits the inclined plane on the ground there. All right, yeah, so that's pretty much it for a solution to part A, because number, we already know the parametric equations, now we just know the interval for T, so when it hits the ground. So that's all essentially what you need it, and we're going to be using that later. Well, actually, I think we are, because I'm solving these problems live as I'm doing this video, so just to save some time. So solutions to part B, all right, so let's take a look at B again. So... Goal the way here. So show that the angle of elevation alpha that will maximize the downhill range is the angle halfway between the plane and the vertical. So basically to maximize this distance here, we're going to, we want to show that it's going to be halfway between this. So half, this angle's half of half. Yeah, so basically half of whatever, the total angle is like that. So that's pretty interesting. So let's take a look at how the solution's manual solves it and see if I can understand it and teach you guys. All right, let's continue further. Wise Instant Racing, there it is. All right, and so let's go all the way to... right here. All right, so solution to B, so the downhill range, that is the distance to the projectile's landing point as measured along the inclined plane is, so what I'll do is, so we're going to have, let's say we're going to launch this setup, well, well, here's the spot we launched, here's the inclined plane, and here's, I'm just going to draw dash line like this. This is going to be our theta, so that's our theta angle. and then this distance right here. Let's say it lands, oh, this is the pointed lens. So you're going to launch it like this with alpha. So that's the angle you're going to launch it with. It's going to go all the way like that, and it's going to go down here. So now this length, this downhill length right there, it's going to be a function of alpha. I'm going to call this R of alpha, like that. All right, that's R of alpha. And now what I'm going to do is I'm going to draw similar triangles. like this, I'm going to put this down here. Yeah, so similar triangles like that. Now, this is parallel, this is going to be a theta as well. So that's theta. And now this horizontal distance, that's the point X right here. Yes, that's just the point X here. So that's number X and that goes all the way down to Y. So that is just over here. Yeah, so that's our X and this goes all the way down to Y. So now this function, And here, quickly just move it to the left and actually rotated a bit flatter. So what we have now is using just similar triangles, or not necessarily similar triangles. Yeah, we're just using the definition of tangent. So tangent, actually not tangent, that's going to be secant. But I'll say cosine, so definition of cosine theta. So it's going to be adjacent over hypotenuse. So this one over this one. So that means we get, this is going to be, this is going to be x over r of alpha, like that. And then just move this over to the other side, and then move this over to this side. What we end up having is, we're going to have, now we're going to get R of alpha is equal to, yeah, x over cosine. Yeah, so we divide cosine theta. and cosine theta, 1 over cosine theta is just going to equal to, well, we're going to have a secant, yeah, but I'll put an x, so we're going to have an x times secant theta, because again, definition ccant is one over cosine, so that's what we have. I'm going to box this whole thing. All right, and here's, we'll write this down for completing, so where x is the coordinate of the landing point calculated in part A. So again, this x just goes down all the way to x here. So that's our x there. And yeah, if we scroll back, so remember there's our x and there's our y equals negative tan theta x. That's where it lands there. That's basically, yeah, what the x component is. All right, and we already know what the x value is that that is just this whole giant thing. So I'm going to do is I'm going to copy this right here. I'm going to copy, actually, this whole thing like that. So that's our x. So we're x coordinate, and then we're going to have R. is going to be R, and then this part right here. All right, now this equals to, I'm going to put the X now that we just had earlier. I'm going to erase this part. Yeah, I'm just going to move this here. So that's our X value. All right, now we're going to multiply this by, well, secant. Yeah, secant right here, secant theta, like that. All right, now going further, now we can do a bunch of simplifications. And first of all, we're going to get rid of these tan and sine and secan, just make them a regular sign and cosine. So we have 2 v0, cosine alpha. Now we're going to have, yeah, this is going to be over G, and now we're going to have a sine, yes, sine alpha over cosine alpha. Alright, and then plus, this is going to be now sine theta over cosine theta, and it's going to be now 1 over cosine theta. All right, now we're going to do is multiply that inside, just for completeness, we'll just do it slow and steady, so 2 v0 squared, cosine, squared alpha, and now we're going to have a, this over G, that's going to be a sine, alpha, cosine, yeah, this is going to be cosine alpha, cosine Theta. Now we're going to have a plus sine squared theta, I'm going to sign sign, it's normal theta, cosine squared theta. So multiply that inside. So we have that. Alright, so now what I'm going to do is, I'm going to factor out the 1 over cosine squared theta. All right, but I'm also going to do is I'm going to factor out the cosine alpha, but just one of them, not the cosine squared. So what we end up having is we're going to have a 2v-not. Now we're going to have a cosine alpha, and now we're going to have a G times it by... uh cosine squared theta so we want the cosine squared theta out of there and then we're gonna get a pretty interesting setup so we're gonna have multiplied now so we brought in so we uh let's just move this in first easy one the cosine alpha moves it in it's gonna be sine alpha and then oh the cosine out we'll only bring one in so this is gonna cancel with this so that's gone all right and then also yeah also we're gonna take out this cosine squared Theta, well, we only have one of them, so we take it out, we need to add another one. This is going to be times by cosine, Theta on top. Again, if you multiply this inside, what ends up happening is you're going to have a, you're going to cancel out this, and then you're going to have a cosine. Theta, exactly this one here. So again, so we're not changing anything. We're just factoring it in and out. So that's gone there. Well, we have, yeah, without, that's gone there. Now the next part is going to be. and it's going to have a sine theta, yeah, so we have sine theta, and then we all multiply the cosine alpha inside, it's cosine alpha, and then we factor out this whole thing all gone, so that's gone. Alright, so now we have this, this is just a, a trigonomic identity that I solved in my earlier video way back, like 10…