In this video, I show that firing a projectile has a maximum total distance traveled in the air when the launch angle is approximately 56°. This is 11° higher than the 45° angle needed to maximize ...
Transcript
the solution to problem 5, this one's going to involve a lot of algebra, and yeah, just writing a whole lot of equations. So, total distance traveled by projectiles. So if a projectile is fired with angle of elevation alpha and initial speed V, then parametric equations for its trajectory are, as for my earlier video on this... on this projectile path and so on, as which I covered also in the earlier problem, so x equals to v cosine alpha, again alpha is the launch angle of elevation alpha times t, and then y is equal to v sine alpha t minus, now because that has gravity affecting it, so minus a gt square. So we know that the range or horizontal distance traveled is maximized when alpha is equal to 45 degrees. Again, I've covered that before. And it's not just 45, but it's also halfway between the, if you have a vertical plane or a downhill plane, it's halfway between the vertical as well. And yeah, so it's pretty amazing. It's always between the vertical, halfway. But now that's to maximize horizontal distance, all right? But that doesn't necessarily mean the total distance travel is the highest. So it's, now we're asked, what value of alpha maximize the total distance, including the upwards one? So what if this one is longer than this one? This one, might be longer because you're firing in the air. So the question is, what value of alpha maximize the total distance travel by the projectile? And it says state your answer correct to the nearest degree. All right, interesting. So let's take a look at the solution. So we know that the trajectory of the projectile is given by, all right, and that's basically combined these into the usual vector equations. So we're going to have the position vector R of T is equal to, in its vector form, v. and then t, and then we're going to have, for the y component, v, sine, alpha, t minus one half, gt squared, like that. Alright, and now the distance that the projectile travels is equal to its arc length, which is equal to the integral of the speed of the particle in time. And you can see that from my screenshot, or just my thumbnail earlier, this one here, so here's the arc length, so notice how it goes up, yeah, that's arc length, and arc length is equal to the integral from A to T, so this A, and then there's T, and this is the magnitude of the derivative of the position vector. In other words, the derivative of the position vector is velocity, and then magnitude of that is the speed, and then it's integrated over time, and you get this, again, just a distance formula there. Yeah, that's just distance in 3D, but this is with an integral, so you're integrating over a whole wide range of the speeds to get the final distance. Yeah, so what that means is we need to get the velocity and then the speed, so thus, So thus what we need is, well, R, you have the velocity, V of T, so v of t is equal to the derivative of the position vector, and now this equals to, well, this is going to be derivative of v cosine, yeah, alpha, and then T, that's just equals to, and then assuming alpha is just constant. that's going to bend in the speed here. Yeah, the speed V is also assuming that's constant, that's the initial velocity. So this is just going to equal to, the derivative is going to be equal to, well, that T is only variable, so v cosine, it equals to one there, so v cosine alpha. And then the Y is going to be well, v sine alpha, and then put the two down, we'll have GT minus GT like that. Alright, so then we have that, and now we go and take the speed, the speed is just going to be, well, the magnitude of V of T, which equals the magnitude of the derivative there, but we'll just, we only know that. So it's going to be now, this whole setup here is going to be now using the distance formula in 3D. So again, the same thing, but we're going to also factor out this V there. So I'm going to take a V out of there. So v squared. take it out, we're going to have now a, actually, I know, actually, never mind, because we have a V, it's a minus GT. So it's going to equal to v squared, cosine squared alpha, plus, now we're going to have this part here, v, sine alpha, yeah, so we can't factor out stuff there. So now we have to actually expand it out, so that's going to be quite interesting. All right, so we have that, and I just extended that. So we're going to have to now foil this out. So it's going to be, it puts all the way to the end, v cosine, a v squared cosine, squared alpha, and then we're going to have this multiplied by itself. So we're going to have a v squared sine alpha square, yeah, sine alpha squared, and then this one multiplied by this. So we're going to have, and then we're going to actually do that twice because when you foil this out. So that one, do that twice, we're going to have a negative, and it's negative. So, negative 2 will have a GV, v, sine, alpha, and then, oh, there's a T as well. And here, quickly, just pause the video, enjoyed it, like, there's 2V, g, sine alpha, and then T, that's all of my, the solution's manual, does it, like that, puts a T on the outside, and then last one we have is just, this negative T, negative GT, minus, multiplied by itself. So we do that, so we're going to get a plus gt squared, so plus gt squared d t squared like that. Alright, so now let's do some simplifications. Well, this part right here, we can just factor out to v squared, and then we're going to have a co-squared alpha plus sine squared alpha. So in this part just equals to 1. Yeah, so that's just the Pythagorean identity. So we do that. Yeah, so this whole setup equals to v squared like that. And now what the calculus book, Solution's manual, does, it factors out, so it's going to factor out a G squared out of all that. So what it does that, yeah, what it does that, yeah, actually, G squared just get rid of that whole setup there. So what it does is equals to, so I'll put it over here, I'll put it, Here, I just move some room there, so put it here, so put this, so we can keep track of it, so magnitude of the speed. I mean, you have the magnitude of the velocity, so you have, which is the speed, this equals two. Yeah, so what's going to do now is move this one in front, and then factor out the G squared. Actually, before we knew that, I'm going to actually, let's write it out first. So I'm going to write this out like this. It puts this first GT squared, so we don't get lost, and then put this negative on it next, after that. 2VG, sine alpha T, and then we put the other one at the end, so v squared like that. So puts it like that, and now what it does is, it factors out this G squared, so we're going to get a, so that we get this G squared out of there. Yeah, so out of this one, it's going to be like that, and then bracket, yeah, we do that, now we're going to have a T squared minus, this one's going to have a G at the bottom, because we have only one G, So we need to, if we're factoring out a G squared, we're going to have to put one divided by it at the bottom. So we're going to have minus 2v, sine alpha, and then divided by G. So again, when you multiply this inside, you're going to, this will cancel, this will cancel, we're just left with the G. So we're not changing anything. We're just doing that, just factoring all that out. All right, and then just put a T there, and also to make it just a bit more fit my solution to my, what it does just make it look a bit nicer. puts this bracket like that and now we have a this here 2 v over g put it like that and then the last one is going to be well v squared we need to divide by g squared yes we're not canceling anything out and yeah here i just box this out yeah so again uh yeah this v squared over g squared and again if you multiply this inside this just cancel so we're just left with this here so again we are not nothing's being changed all right let's close uh this stuff so now let's keep simplifying And yeah, we could see that, and the reason that, yeah, the solution is that this is now we can complete the square. So we can go, I'll just write C, let's write, complete the square, as we did before, complete the square. Yeah, so this is just, you know, getting the, the quadratic setup, and also we can factor out this G out of there, out of the... the square root completely. Alright, if I just now scroll back up, all right here you get quickly teleported above, I think it's problem one over there. So now there's the AX square plus B, X plus C, complete the square, and then we can get this setup like that. It's basically move this into a square there, where, yeah, where we have D, X plus D is equal to, well, B divided by 2A, where B is this, and then A is this one here. So we'll take a look at our case. And yeah, so go teleport back here. So we're going to get now, this is going to be, the setup's going to be T. I'll just do the complete of the square one. T minus, well, it's actually a T plus, yeah, T plus this, this is our B divided by 2A.A is one. So we're going to get 2 minus, and then we're going to have 2 v, yeah, over G, sine. Yeah, let's put a sine alpha. Yeah, we don't have that other bracket there. So, sine alpha squared. And again, you could expend this out, see what happens. So, again, and it also divided by 2, so we cancel this out. So this is gone. So, and again, you could just see what happens here. Now, if we factor, expend us out, we're going to get a t squared. So multiply this by this one twice. So we're going to get negative. minus 2 v over g uh yeah uh this is gonna be sine alpha and then square and then t and then lastly this multiplied by itself so we're gonna get plus v squared over g over uh yeah g squared move some room there then sign squared alpha like that So that's what this whole square equal. And yeah, actually, instead of this last part here, so we can ignore that part. This is basically completing the square here, this two. And then this one is just, yeah, separate. But now notice what we have, we have this, is this whole setup. And then, so in other words, we just need to subtract by this one. So move this over to the other side, we get this, which is this. And, yeah, so what we end up having is, yeah, so that's how, yeah, completely square. So thus, Yeah, we're going to get thus. This goes down, we're going to get the magnitude of the velocity. Again, I totally is going to be a lot of algebra. So this factors out, so we just factor out of the, or just square root, I mean, we just square root it. This is going to be, well, loss or minus, G, we'll just take the positive size, we only want magnitude, and then square root, and now we're going to have this set up here. Yeah, T minus V over G, g, sign. alpha, and then all squared, like that. So now we only have sign once. This is going to be now, actually, we have it twice there. And this is going to equal, actually, we got rid of the, this, yeah, so just so we don't have a T, a T, a T squared and a T. That's why we complete the square. Now we're going to have this plus V, G, a V squared over G squared, minus now this part. So that, because that's what happens when you squares, we got to subtract that. v squared over g squared sine squared g squared alpha. Yeah, sine squared alpha. Like that, so that's a pretty epic setup there. Alright, now going further, again, there's another ingenious setup by solutions by any of this. So it looks at this one here, if you factor out this part, the VG squared, you're going to get a 1 minus sine squared alpha. And again, if we recall our Pythagorean identity, Yeah, that is going to be now cosine, yeah, this is going to be, yeah, this is going to be just equals to cosine squared alpha. Again, that's from the Pythagorean identity, so we have sine squared, yeah, sine squared alpha, yeah, I'll just say sine squared theta, generic, theta plus co-squared theta equals to one, and then move this over to the other side, and we're just going to get, uh yeah, move it over to the other side, we get a cosine, co-squared, theta, yeah, I mean, one minus sine squared, theta equals to co-square. Yeah, so that's all we get, so just do that. Uh, yeah, I'll just, I'll keep it as an alpha, just for, um, the identity, because it's, we're throwing that in there. And that's, yeah, that's Pythagran identity. So again, which is extremely, extremely useful. So, and then, uh, yeah, so now we have, uh, that we're gonna get now V, I'll put it like this. Yeah, so finally we get this set up every time. Yeah, so the magnitude of velocity or the speed is going to equal to G, and then, you just move it down even further, so have some room, so this is going to be G, and then T minus V over G, sine alpha, squared, and then that's just going to be now, yeah, this part here, so it's just going to equal to v squared, plus v squared over g squared, cosine alpha. Yeah, cosine squared alpha. All right, now that we finally have this, so that's our speed, and now the projectile hits the ground when, because that's what we want to take the integral of that to get the arc length. So remember, it's the arc length, we want to take the integral of it from zero to, well, at the point it hits the ground, so the total distance. So to find when it hits the ground, we have to set this equal to zero. So I'll just take this. All right, here, I'll take that, yes, basically, when that equals zero, so projector hits the ground when the horizontal, the vertical distance, I mean, the vertical component equals to zero, so when the Y component equals to zero, when it hits the ground, so equals zero, and then again, this equals zero when, that's the generic case, when T equals zero, that's the beginning. So we're just launching that, and then basically solve this, we're going to get, well, again, it's going to cancel all the T's out. cancel the T's out, and so on, and move this over to the other side. Yeah, move this over to the other side, we're going to get, and cancel the T, we're going to get, and then move that in the front. We're going to have one-half G-T, so we cancel all the T already, and then this is going to be v, sine, alpha, and then, again, divide these out. So divide, divide, divide both out. So we're going to have the two on top, and the G in the bottom. So we're going to get, t is equal to is the impact time g is equal to uh 2 v sign alpha over g so that's the impact time all right here's box that in yes so uh the distance travel by the projectile is again using just the arc length uh set up there so this is going to be i'm going to call this l as a function of alpha so this equals to because we're taking the full integral and then the only thing that it matters is that alpha is going to be changing. So this is going to be two, yes, so integral from 0 to 2 v over, yeah, V over G sine alpha. Yeah, sine alpha like that. Yeah, so that's our T value, and now this is going to equal to, well, the magnitude of the velocity, which gets the speed, and then D, DT, so DT, so. So again, and then distance is just the velocity, yeah, times, the speed times by time. So how much distance is covered is basically at this rate you're going, for example, meters per second or feet per second, times by seconds, you can cancel you get the unit of distance, which is M, or meters or feet, etc. All right, so now this equals to, we're going to get a ginormous equation, and it's going to be, yeah, I'm not sure, I'm going to solve all this, it's going to be now, put this again, I'll put it again as a 2V sign alpha per G. I don't know why it puts the bracket there. That's the solution's manual of this, though. I'm just going to leave it like this. This is a bit easier for my sake. And then this is going to be G, and then we're going to plug in this full giant thing there. So now we're going to have the G there. So, square root, the whole setup there. and then DT, yeah, we'll put that after, so T minus V over G, so we get T minus, actually, yeah, V over G, sine alpha, T, yeah, squared, and then plus v squared over G squared, all right, plus v squared over G squared, over G squared, and now we're going to get a cosine squared alpha over alpha, and then this is going to be, this is going to be D, DT. All right, now, this integral is of the form in Formula 21 in the table of integrals at the end of my calculus book, which I also did an example on using it in my early video, and again, yeah, there's not a chance in the world I'm going to do this, so this manually is going to take several. pages but yeah i'm pretty sure you're gonna use a substitution and or integration by part and so on but it's of this form right here so notice how we have this squared there and this other part we have this squared everywhere as well so that's also squared and then there's a square root there so uh recall from our other video um i mean not recall from it this is my other videos of hive notes now here's a table 21 in the calculus book uh at the end of the chapter i mean at the end of the calculus book at the end and there's a table, so formula 21. So forms involving square root a squared plus u squared, where a is greater than zero, keep that in mind. So it helps us pick which one of these is A and which one's U. So we have square root a squared plus u, a integral of that equals to u over two, and then times by square root a square plus u squared, and then plus a squared, divided by two, lawn u plus square root a squared plus u squared, and then plus the integration constant. So in our case, so, right here, so in our case, we have, well, this one right here, our A is going to be actually, yeah, this is going to be, yeah, this is going to be equals to v over G, cosine alpha, yeah, that's this part here, and that's because, This right here is greater than 0 for alpha is again between our launch angle. We're not going to be launching backwards. We're not going to be launching in the ground. Yeah, so we're just going to be an actual normal launch. So normal launch is like that. So there's our alpha. And our alpha is going to be between pi or two radiance or 90 degrees and greater than zero. So we're not actually, yeah, so it can't be equal to zero. So this is going to be greater than zero. And just because this one here, we have a T-minus, and yeah, this could be 0 inside this, we're not 100% sure. So I'll put that there, this is going to be greater than 0, like that. And again, yeah, I just fixed that up, but also recall the cosine function, so cosine theta, if you have theta like this, it's going to go down and up, yeah. And this is going to be at 0, this is pi over 2. Yeah, so this is going to be plus, so we're not going to have a negative there. So that's set up. And also, yeah, so where, in our case, A is that, and, yeah, so then U, I'll put, and, U is equal to, just going to be equal to this part, T minus V over G, sine theta. I'm sine alpha, V over G, sine, alpha, like that. So that's our U, there's our A, there's our U, and this G, we just take it out, just the constant. And, yeah, I'm just going to box this, so it's easier to see, so that and that's this. So thus we get, just plug this all inside our formula. So our length, this is the length in terms of, yeah, in terms of alpha, so thus L alpha, put it all the way here. is equal to the integral. Actually, no, we already did the integral. So now we're going to evaluate it. So we're going to equal to this, but we're going to evaluate it over 0 to our impact time, so T. So it's going to be all this equals to. So we'll factor out the G. Yes, we're going to factor out this G there. So we're going to factor out this G. Actually, we factor out before we even do the bracket. So it's going to be G. bracket, all right, so the next setup is going to be U, so put a U there, divide by 2. So we're going to get a U is going to be T minus V over G, sine alpha, divided by 2. All right, so we got that. Now the next one is going to be, well, square root, and then A squared plus…