Example 5: Horizontal Distance Traveled by a Projectile by @mes on 3Speak
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Example 5, and that's the horizontal distance of a projectile. So a projectile is fired with angle of elevation alpha, and initial velocity V0, as in the figure below. So again, there's the initial angle, and it's launching it out like that, and it's going to go in a parabola setup like that. All right, and then there's distance D there. So assuming that air resistance is negligible, and the only external force is due to gravity, find the position function r of t of the projectile. All right, and then now we're asked, what value of alpha maximizes the range or the horizontal distance traveled? All right, yes, we got to basically see what's the maximum distance D there for an angle, what angle gives the maximum distance. All right, so let's go further. So we set up, let's look at the solution, so we set up the axes. or the axis so that the projectile starts to the origin again. That's just, as in this figure, so we start at the origin there. So since the force due to gravity acts downwards, we have, yeah, what we end up having is, again, it's only force acting on it after you launch it, and that's because air resistance is negligible in this case. So what we'll have is the force due to gravity is going to be F like that of T, and this equals to, well, M times acceleration. and that is going to be, well, this equals to negative M, G, and yeah, that's just negative MG, because acceleration A is just gravity, and it's going to be negative, yeah, negative G, so it's aiming downwards, and that is going to be, well, in the Y components or the J component, so put that there, in the J component. All right, and then we're going to say where, and move it over here, so where, Yeah, where G is equal to the magnitude of the acceleration of gravity there, and this is approximately equal to 9.8 meters per second, or 9.81 p meters per second. Let's leave it at 9.8. Round that up. So 9.8 meters per second. Yeah, meters per second squared. Yeah, so that's what it is. So basically, it's meters per second per second. So it's basically velocity. is increasing by 9.8 meters per second every second. Yeah, hence why the velocity is increasing, yeah, so that's basically how you think about it. So if velocity is increasing by 9.8 meters per second, every second, after two seconds are going to be increasing, it's going to be double the speed himself. Alright, so thus, what we have is, yeah, so thus the acceleration, the acceleration is going to equal to, well, just negative G, so we just move that M out of there. So we can just cancel these, just get the negative G. So acceleration is this equal to negative G, J. And now again, we're asked to find the distance D. In other words, we basically just need the position vector to get this all. So we get the velocity, then we get to the position. So we just take the integral. All right, so since the derivative of the velocity, V of T, is equal to A, then Let's put, we have, so what we have is now, well, the integral now, we just take the integral of it now. So, yeah, so V of T equals to the integral of a, of, yeah, DT, like that, and this equals to, well, integral from, yeah, this is going to be negative G. Let's put this in a bracket, actually. negative g like that d t but also yeah it's a vectors we got to the j component there negative g j then d t like that so this equals to yeah and now the integral of negative g that's just going to be well a negative g t j then we have to add a constant this is going to be plus a constant c like that Yeah, where C is just going to equal to, as equals to v0 at t equals 0. Because you could just put in, yeah, V at T equals 0. Because you could just plug in v0 here, and then you're going to get, this is going to vanish, or you just love with C. So that's just equals to, yeah, I'll call this v0. And again, this is, as I went over earlier, and I basically showed that. in general that's what we have whenever we have uh yeah in general of vector intervals also recover velocity when acceleration is known and it equals like this so v t not um yeah the starting position we just made it at t equals here and then this is um Yeah, and then the position one, when you know velocity, you just go to R, R, T, not, and so, or R0, this is V0 and R0, and so, it's fascinating stuff, and then plus the integral of that acceleration, and so on, all right, let's continue further. All right, let's go back here, so thus, what we have is, V of T is equal to negative G, T, J, Yeah, JTJ, and then plus v0 vector, like that. All right, and now integrating again, we can get the position vector. So we're going to get a, we're just going to get, I'll just put it like this, R of T, now is equal to, well, the integral of the velocity, DT, and now this equals to, well, integral from negative tj, and then plus v, not like that. I'll put a bracket, v, not like that, this is d t. Now this equals to, well, integral of negative gt, that's going to be, well, gt squared, and then divided by two. Because you have to, when you, for example, when you take the derivative, put the two down, those cancel and so on. This is J still, and now plus v0 times T, and now plus D, like that, to put another position vector, I mean, I mean the integration constant, and where, where, likewise, D right here, is this equal to, well, R of zero there. because we're starting at the origin, we're starting from there, and then put the zero inside, this all just vanishes. And so on. Yeah, it vanishes, and since we start at the origin, this is just equals to the zero vector or the origin. It's just, we just add to nothing. So that's the starting point there. So there it is. So we just have it like that, so we just remove it all. So thus, I'll just type it out, actually. So thus, the position vector of the projectile is given by, and I'll move this a bit higher, actually. A bit higher. So, thus the position vector of the projectile is, and as the just vanishes, so we get a R of T, because now we need to maximize this value. That's basically the, or, yeah, just the full length of it. Well, actually, we need to maximize the horizontal distance of it, and we're going to break that into component. So what we'll do is I'm going to write this down. So negative gt squared over... I'll put it over 2, J, and now plus the calculus book puts a T in front, so T, then V vector not like that, so that's what we have, so that's a position vector. All right, well, actually, yeah, I'm not going to bowl this one, because this isn't the final thing. We need to maximize the distance, that's the thing, that's the key part, so let's keep going further. All right, going further, so now, if we write that the magnitude of the initial velocity vector, it's equal to just v-0-V -not-v-that, so not a vector, it's a scalar, just put it like that, just put the italics v there. The initial speed of the projectile, then what we have is, well, if we look at our... set up here, I'm going to write this out like this. So we have our X, this is our Y. So we have vector. This is our starting vector here. This is our alpha. And now we're going to get, this is our V-0. So this is our V-0 equals to V-0, like that. Actually, I'm just going to put V-0, yeah. We already wrote it over here. So we already wrote it there. So that's our V-0. like that, that's the length, and this is going to be our vector of v, not like vector like that. And now, this is, draw our typical trigonomic triangle, as always. I just fix that up, so we have it like this. So what we get now is what I'll call this, I'm going to call this V, not the Y component. And now this part here, again, there's the origin. All right, and now this, the adjacent side I'm going to make this V0X, or V0X. All right, so again, using basic trigonometry, what we have as well, we know a sine alpha is equal to, well, it's going to be opposite v0y component over the hypotenus, which is just the magnitude, the V0. Like that. and then move it over to the other side like that. This is going to equal to v.0, sine alpha. All right, and then likewise, cosine of alpha is equal to adjacent over that. So v.0x over v0, move this over to the other side. What we get now, this equals to v, yeah, v.0, just the magnitude of that initial velocity. v0, and then times it by v, I mean, cosine, and then this is going to be omega, I mean alpha. And here I quickly just fix this up, and I added this just for completeness, so v0, the magnitude vector of that equals v0, yeah, so v0 vector there. So thus, what we get is v0 is equal to So we're going to put this in its components. Yeah, just because we have this right here. So we're going to put this in its components. So this is going to equal to v .0 cosine alpha, i, plus v0, and then this is going to be sine alpha, like that. All right, so thus the position vector of the projectile become, so we have this whole setup here. I'll just copy this, instead of writing it down again, I'll copy this, and paste it in here. So this becomes now, we just plug in this V-0 inside there. So we plug in this whole thing inside here. All right, so we're going to get this equals to, yeah, this equals two, well, this one has an I component, and this one doesn't have an I component, it's just a J, so that means, We're going to have this part here. So we're going to have, here, this right here, the I component, I'm going to write this out. I'll put in a bracket. So v0, cosine, alpha, but then we're going to multiply by T. It's multiplied by T. So we put the bracket so it doesn't look like it's alpha T inside there. Now we have the I component. And now, plus, now the next one is going to be, well, this is going to be inside here. So what that will look like is this. This will look like all the two brackets. It's going to be like a bracket. Yeah, square bracket and then the regular bracket, this is going to be now, we're going to have this is the positive one, and put the negative on the right. So v0, sine theta, v0, sine, I mean, sine alpha, and then times it by T, because we're multiplying this inside here, this T, and now we're going to have to have subtracted by GT squared, GT, yeah, squared over 2. Yeah, and this is going to be a closed bracket. and this is our, this is our J, factor like that. All right, so now that we have this, the next setup is, well, we can have now our parametric equations of the trajectory are therefore, and basically take the X and Y, come on, so this is the X one, so the X component is equal to, there's just the I, I set up there, so this is going to be now equals to v0, cosine alpha times T. and then the y component is going to be equal to v0, sine alpha times t minus gt squared over 2 like that. All right, now we have all this. Now this is what we box in, because this is pretty much what we can start to obtain a distance, because we want to get the maximum distance, horizontal distance, traveled. for basically the alpha, which alpha gives the highest, the farthest distance? So what value maximize the range of horizontal distance travels, basically maximizes D? So that would need to find the perfect angle. It should be 45 degrees, that's my guess. Alright, and now just a quick note here, just a side quest here. So if you eliminate T from the above equations, you'll see that Y is a quadratic function of X. So the path of the projectile is part of a parabola. So for example, if I just take this right here and then move it over to over there, so we have T as a function of x. So what we get is t equals to x over v0 cosine alpha. And now we throw this inside the y there. So what we get here is we get y. is equal to, well, we got a v0. Yeah, we just throw the T inside there, so I'm going to put this like this. We have this v, uh, not sine alpha, then times by t, now it's going to be times by x over v, this is cosine alpha. Like that. Here, just fix that up, I put the alpha at the bottom there. Yeah, so that's the alpha. It's hard to write alpha. All right, we'll do one more time. Alright, so there's alpha. And then, yeah, so v.0, sine alpha divided by v.0 cosine alpha, and then minus, this is going to be, well, G, and now the T is going to be this setup here, so G x squared over, well, there's a 2 here, so it's going to be over 2 v0 squared cosine alpha. squared like that. Yes, in other words, well, this cancels, and then we're going to get a, the top part's going to be, well, 10, 10 alpha. So it's cancel like that, like that, to make it thinner so we can see it. So this equals 2, so yeah, we get y is equal to 10 alpha x, and then minus, again, this whole set up there, so G. X squared over, I'll put it over here, let's move it over, actually, erase that, move it here, put this in a bracket, so it doesn't look like it's part of the tan, and then minus, move that all these giant constants by itself, so G2v, not squared, cosine, alpha, squared, and then times it by x squared. All right, so yeah, so this is just a quadratic formula, this is just a problem. I'll just put this like this. I'm not going to graph this out, because we need all the other variables and so on, but it's just a problem, because it's just the x squared, there's an x, and you just may plate that, and everything else is the concept, so it's just a problem. All right, so continuing further and again, this problem, whenever you have an x squared, x, or just everything else is constants. You just get a problem, one of the highest powers, x squared, right? So going further, so the horizontal distance D is the value of X when Y is equal to 0. Yeah, so basically, we have the parabola, so it's basically parable, so we have the starting point, Y equals 0, so basically when Y hits the 0 again, that's going to be the distance D, so at that X value. So let's continue further. So setting Y equals 0, we obtain, yeah, so what we obtain, well, we're going to get a Y, a T value, and then we're going to throw that inside the X one. So we have this right here. So y equals to this whole set up there. So y equals to v0, sine alpha t. Yes, let's go back up here, so, and then, yeah, sine alpha t, minus gt squared, minus gt squared over two. and now set this equal to zero, so that's when it's equal to zero, so now, and then we can basically cancel out one of the T's, or instead of that, I'll just factor it out, so what we're going to get is a, we're going to get a T, and then, yeah, so that's T times it by, then we have a V -0, sine alpha minus GT over two, this equals to zero. So we have two setups we have the, at the origin, this is at equal zero, at t equals to zero, and equal zero, or this whole setup is, yeah, I just put it like this. So we have this, this can equal zero, or this whole thing can equal zero, so that this whole thing becomes zero. All right. Yeah, so thus what we have is, thus, t is equal to zero, or that's the origin, that's a trivial one, or we have or when then, if this is equal to zero, so then these two are equal to each other, so, or this equals to the negative. This one, yeah, move it over to other side, so they're equal to each other. So we're going to get a, you know, we're basically set it equal to each other, and then move these over to the other side. So what we get is, whoops, G. So what we get is equals to t is equal to, move this over the other side, v0, sine alpha, and then move the T2 on the other side, so two times it by that and then divided by G. Like that. So I want to box these in, and those are the answers for T. Now we just plug that into our X value, and we'll find the maximum. All right, so now this second value of t gives, and now we'll plug it in, so d is equal to the distance, is equal to, well, just the x component, and then that is going to be equal to, uh, where is the parametric value of v, not, cosine, alpha, then times it by t. So v, not, yeah, right down here, v, not. All right, and then, uh, cosine. alpha and then t. I'll put the whole thing in. Yeah, so again, that's v0, cosine alpha, t, and then throw this whole thing inside there. All right, so throw that inside. All right, what we're going to get is this equals to a v0, cosine alpha, alpha, and then times it by, I'll put a bracket, times it by 2v0, sine alpha, oops, divided by g. And simplifying this, well, this equals 2, and then the v, we'll have a 2, I'll put the v0 squared first, so v0 squared, those ones squared, divided by g, and then we're going to have a multiplication here times by 2, sine alpha, cosine alpha. So this is going to be a 2, and the reason I put it like this is, because this is just, we already know the identity for this, is going to be sine alpha, and then cosine alpha, and the identity, which I covered, proved in my earlier video, put the alpha like this, this right here equals to just sine 2 alpha. Yeah, that's what equals. Yeah, and I'll link back to my earlier video on that. Basically, now this whole thing becomes now, well, D is equal to x, and that that equals to v0, I'll put it like this, v0 squared, sine 2 alpha, divided by g, and that is our distance formula. And now we need to know when it's maximum, and that's going to be... Yeah, it's going to be when sine 2 alpha's maximum. In other words, typically it's a, you'll have pi for 2, but there's a 2 there, so divide by 2, it's going to be pi per 4, and I'll show you that in a bit. But first, recall the sign double angle identity for my earlier video, and then I just took a screenshot from my calculus book. So this is the double angle, sine 2x equals to 2 sine x, and there's the other ones for a cosine, and then for tangent as well. And this one has multiple ones. and so on i got you can end this one here anyways so you can see a link to my earlier video to proof on this one or let's go further so clearly d has its maximum value when sine 2 alpha is equal to 1 that is when alpha is equal to pi over 4 radiance. And you can see that because, again, it's a sine function, so this is the only thing we have here. So this one's the constant squared, and then the G is a constant. So the only thing that changes is the sine 2 alpha. So just recall the graph for sine theta. Yeah, just fix that up. So that's theta, and there's sine theta. So then it looks like this, and goes up and down, and so on. I'll fix this up a bit like that. Alright, now here's the origin, and then add here is 1, then here is negative 1, and then at here is going to be pi over 2. That's a maximum there, but that is when theta is equal to pi over 2. Or better yet, I just move it over there, and I'll just say maximum at... at theta equals pi per 2. So that's the maximum at theta equals piver 2, where in this case, theta, because we have sine theta, it's going to be sine 2 alpha. So it's going to be, well, equals to 2 alpha. Whoops, this is going to be like that. So in other words, move this over, we get alpha is equal to pi over 4.