Problems Plus 1: Projectile fired from the origin by @mes on 3Speak
In this video, I examine projectile motion launched from the origin and show that the projectile will always lie within a parabola defined by launching a projectile from the maximum height horizont...
Tags: #math #calculus #science #education #projectilemotion #threespeak #vectors #chessbrothers #scifimultiverse
solution to problem 1. So, projectile fired from origin. So a projectile is fired from the origin with angle of elevation alpha and initial speed V-0. Assuming that irresistance is negligible and that the only force acting on the projectile is gravity, G, we showed in my earlier video that the position vector of the projectile is, so R-FT equals V-0. cosine, alpha, and then t, and then i vector, plus v0, sine alpha t minus one half gt squared, and then all times, and this is all times by the J standard basis vector. We also show that the maximum horizontal distance or range of the projectile is achieved when alpha is equal to 45 degrees, and in this case the ranges r equals to v0 squared divided by g. And again, I covered this in my earlier. video there is linked to the hive notes, sir. And then part A says, at what angle should the projector be fired to achieve maximum height, and what is the maximum height? And then we're asked, fix the initial speed v0, and consider the parable x squared plus 2 r, y minus r squared equal 0. And that's this pink one right here. And again, as I stated earlier, if you made x equals 0, this vanishes, and then r if you make it equals to r or over 2, this 2 is cancel, we get an r squared, r squared minus r is equal to 0. So in other words, this is going to be one half of R. Like that, that's the total height of this dome there. All right, so if we have that, and then we're asked to show that the projectile can hit any target inside or on the boundary, so it can fire like this inside or right on it, etc., but it can't hit anything outside of it. or on the boundary of the region bounded by the parabola and the x-axis, and that it can't hit any target outside, so it cannot hit past it. That's because this V-0, it's basically we're setting it equal to this V-0. I'm pretty sure, yeah, so it is. Yeah, that's, because we're going to be fixed initial speed, V-0, and so on. All right. Yeah, and that's basically you're firing initial speed and it's going to go down with gravity. So you cannot shoot past that from the origin. So, and then see, suppose that the gun is elevated at an angle of inclination alpha in order to aim at a target that is suspended at a height H directly over point D units downward. yet downrange, target is released at the instant the gun is fired, show that the projector always hits the target, regardless of the value V-not provided the projector has not hit the ground before D. So you have H, if you drop it, a distance D, so there's going to be a relation between these, that doesn't matter what speed that you're launching this out, it's going to hit this exact spot, assuming that this height H doesn't go through the ground, doesn't hit the ground before D, all right? So going further, so now let's take a look at the solution to Part A. Alright, so let's scroll up to A again. So A, we're asked at what angle should the projectile be fired to achieve the maximum height, and what is the maximum height. So we've got to find out what the angle is for the maximum height, and the maximum height itself. So I'm going to copy this. So that is our position vector of the projectile, and we're going to be launching it from the origin. It's going to be like that. So we're going to be launching it. So I'm going to copy and paste this. So this is, again, from my earlier video, I showed that this is the case of a projectile. So the projectile reaches maximum height when, and then basically, so that's the position vector. And then if you have the x, y, axis, just for completeness, X, Y. So we're going to be launching it, and it's going to be like this. It's going to be launching it, it's going to land like here. So the maximum height, here is the maximum height, and that is at, this is going to be at the derivative Y prime is equal to 0. So basically, DY over DT is equal to 0. So this is flat. So that derivative is going to be 0 there. And this is going to be, this is V0. Yeah, this is output of V0 like that. That's the speed there, initial speed, and there's the angle alpha. All right, so at y prime, again, that just equals to DY over DT, over DT equals to 0, and that equals to, well, we're going to be taking the derivative of the Y component, this is the Y component, that's the Y equals that, so that's not just going to be, well, D over DT of this whole segment there, V-0, sine alpha. t minus one half gt squared. Alright, so yeah, that's all we have to do, actually, I'm just going to put it like this. Alright, so if we take a derivative, this just equals, so we're going to have a zero equals two. Well, derivative of this part here, that's just, uh, t, so that's just derivative is just this right here, v, not sine theta, I'm sign alpha, v, not sine alpha. and then minus, this is a two, so take the two down, so two times two over two, it's can cancel, and then minus the power, so GT, so that's without it. All right, so we have this, now we just solve for T, so we could just move this over to the other side, solve for T, we're going to get a, well, GT equals that, divide by T out, I'm divided by G, so we're going to have equals to, G is equal to v0, sine alpha over G, so this is, T when at the maximum height. So this is the time of the projectile when it reaches the maximum height. All right, so now we have this. So what we're going to do is, well, we have the Y value is this. That's the Y component. So we have the maximum T value. We just plug that inside. So we have Y. So let's put a thus. Thus. Let's put it for completeness. So we have y equals to v0 sine alpha, v0, sine alpha, and then this is going to be T, so we're going to be throwing this inside there, minus G, uh, yeah, one-half gt squared. So this is going to be one-half g-t squared. So this is going to be one-half g-t squared. I'm going to throw that in there and in there. Yes, so this equals 2, so v0, sine alpha. Fix this S, all right. I'm going to just fix that there, so then now we're going to put the T inside, that's going to be, well, V0, sine alpha over G, and then minus one half, and now we're going to have a G, and then that whole setup there. So we're going to have a v0, sine alpha, and then this is going to be over G, and then squared, like that. All right, now that we have this, we could start doing some simplifications, or just expand it all out. So we have V0 here, this is a V0, so that just becomes, well, v0 squared. Next one, sine, sine alpha, that's going to be, well, sine alpha, that's going to be, well, sine alpha squared. and then divided by G. Now the next part is going to be one half, and now this G, there's me a G squared, so then one of the Gs cancels, so we just out left of it divided by G. This one's v0 squared, sine squared, sine squared alpha. So we have v.0 squared, sine squared alpha, all right, and then divided by G. And it's exactly the same thing on the other side, but this is a half fit, so a full divided about, minus a half, just a half. So half, you could also do 2 over 2, but I'm just going to know that. So what we have is, yeah, what we have is, well, this minus half, it just equals a half. So what we're going to get is Y, this is going to be the maximum height, Y equals 2, I'll put this all together, so v, uh, uh, not squared, yeah, so v, not squared, sine, uh, yeah, squared alpha, and then divided by, uh, 2G. So, put the G and put it, too like that, so that's what we have. All right here, I just fix up that box, and now this is the maximum height obtained when the projectile was fired with an angle of elevation alpha, and this maximum height is largest when alpha is equal to 90 degrees, and which makes sense as, yeah, as it means the launch angle is straight up vertically, yeah, basically we're just launching it straight in the air, so this is going to be a 90 degrees. So we're launching that projectile, or just object. Yeah, we're just going to be launching whatever it is, bullet, ball, et cetera. So we're launching it in the air like that, and also, again, we know, recall that the trigonomic sign function. Just recall it looks like this, and this is going to be at the maximum height is going to be 1. This is going to be at sine theta, and this is at piver 2, radiance, here's the origin. All right, and now this point right here, this is going to be, yeah, at this height, let's erase this, at this height, this is going to be at, that theta is equal to, yeah, pi or two radiance, or that's just 90 degrees. All right, yeah, so that's the height, and then what we get is that equals to 1, and then sine squared of 1. Yeah, 1 squared is going to be also 1. So we're just going to be locked with the maximum height v squared, v0 squared over 2G. And yeah, so I wrote that right here. So in that case, sine alpha is equal to 1, and the maximum height is, is v0 squared over 2G. I'm going to write this for completeness. So we have this right here, y equals v0 squared, sine squared alpha over 2g. So I'm going to write y equals 2, and I'll write this v0 squared over 2g. and then just put this sine squared alpha. Yeah, so then this is going to be well equals to 1. So then this vanishes, we're just left with v0 over 2G. Yeah, so perfect stuff there. Alright, and now let's take a look at the solution to part B, so let's read it up again. So part B states... fix the initial speed v0 and consider the parabola x squared plus 2 r y minus r squared equal 0 whose graph is shown in the figure below that's basically this graph right there so let's go and write this down so that's x squared plus 2 r y minus r squared all right so we are asked to consider the parabola here so that's probably just copy and paste to just make it easier so we have yeah this this this parabola right here this is going to be our x squared and then it's going to be plus 2RY minus r squared equals to 0. Yeah, so you can see right here, yeah, you can see right here at x equals to 0, so I'll go at, yeah, at x equals to 0, what we get is, well, we get 2RY minus r squared equals to 0, and then basically, yeah, so y has to be, to make it equal 0, so y, is equal to r over 2. Yeah, so then when y equals r over 2, we get basically, we get 2, r squared over 2. This cancels minus r squared equals 0. So that that cancels. So basically, this height right here, this right here, if you're launching straight up, that's going to be r divided by 2. Yeah, so what I'll do is I'll put this arrow at the general function, and then I'll put this is going to be at r over 2. And that's going to be the maximum height if you just launch it straight up. Yeah, so then, so, thus, let R equals 2, well, we take the maximum height right here, and then times by 2, because the numbers are divided by 2. So we just take this maximum height, v0 squared over 2G, but we multiply by 2 because the maximum height, is that, so then basically, yeah, we'll just put it like this, thus, I'll put it, the maximum height, thus max height, is if you launch straight up, is equal to R over 2, equals 2, and then this is going to be v0 squared, v0 squared over 2G. Yeah, so then basically cancel the 2s out, so we get R is equal to, and this is just going to be, well, V0 squared over G. So we got rid of the 2 there. All right, and now what we could do is, well, we could rearrange the parabola as an equation for y. So we have this right here, just x squared plus 2 r y minus r squared equal 0. So we rearrange that, and then just solve for y, so we have x squared plus 2 r y minus R squared equals to 0, so let's just move this things around. Move this over to this side, move this over to the other side, so we're going to get a 2RY equals to, and my calculus book puts the x squared first, so we can put a negative x squared plus r, I mean, R squared, and now we just divide by 2R on both sides, so we're going to get a, move it over to the other side, as we get y is equal to, well, negative, Yeah, negative, I'll put that negative better like this. Alright, this can be negative x squared, but this is better, x squared over 2r, and then plus, well, one of the r's cancel, we're going to be left with r over 2. So that is what our y, I mean the problem as a function of y. Yeah, I mean as a function of x for y, so basically y of x equals to this setup there. Alright, and now the points on or inside this parabola are those for which, and basically at the domain ranges, we have this negative R, there's the R, it goes up to R 1.5. I'm going to draw this out just for completeness there. So we have this setup like this. So we have, this is our parabola. This is going to be our Y, that's our X, on the X, x, just like that. This is the origin, so we have the parabola looks like that, and this height is going to be R over 2, that's at this point. like that, whoops, I made a mistake, I don't know what happened there, so I'll just undo, undo, okay, there it is, so erase that, so that's R, and then at this point right here, these two, that's going to be negative R, that's going to be R, and why is it doing that? Yeah, I don't know why it's doing that, but anyways, well, hopefully it doesn't do that again, so now we're going to have, this is going to be our y, so that's the problem, y equals to negative x squared over 2r, plus R over 2. Yeah, so the domain is going to be well, basically from negative R to R, so X is going to be greater than or equal to negative R, and less than or equal to, and this is where all the points are inside, on or inside here, so the whole thing within there. I'll actually, I will do that bold bit, actually. Yeah, I will fill it in, actually. Yeah, I'll just fill that in again here. So I'm going to fill this in. Like that, yeah, so basically we want everything inside, and so on. So, and then x is going to be less than r, like that. And we have y, well, it's going to be greater than a 0, so y equals 0 greater than that. So, 0 is less than or equal to y, and this is going to be less than or equal to, well, just the parabola x squared over 2 r plus r over 2. Alright, and now before we go further, I'm going to quickly recall from my earlier videos on the formula for the position of a projectile. And again, recall my earlier videos, the hive notes there. And then what else I've done is the screenshots of the sections videos there. So you can scroll down to the bottom, you'll see these thumbnails, and so on. Basically, there's the projectile launched. You get this position vector. So R equals to v, cosine, alpha, and then I, then that's the X of 1 .1 is the J, or the Y common of v, not sine alpha, and then minus 1 half GT squared. And then also for the example I did right here, you could use the parametric equations, or x equals v.0 cosine alpha T, and this is the y, y equals v .0, sine, alpha, t, minus 1.5, gt squared. And again, we were already given this in the question, I mean, in the problem. It says from the early video, that's the same when I just showed the thumbnails, and there's that setup there. Alright, so we have that quick recap, let's continue further now. And yeah, also one more thing in my recap here, so for the Y component, when y equals to zero, so when it besides the origin, it goes all the way up to here, so there's one Y equals to zero, we, I showed my earlier video that T is equal to 2v0 sine alpha over G, so that's the Y value when it hits the ground there. So when the projector is fired at an angle of elevation alpha, the points X, Y, along its path satisfies the relations. We write the X1, that's just going to be, well, x equals 2, v0, and then v0, uh, uh, cosine alpha, cosine alpha t, and y is equal to this is v .0, sine alpha t minus one half g to square. v0, sine alpha, and then t minus one half gt squared, and then this is where, yeah, this is where we have, well, zero is less than equal to t, so t is between that, and then obviously this part right here, so we're, t is equal to 2 when the object hits the ground again 2 v0 sign alpha over g so this is 2 v not sign alpha over g actually is there the 2v not sign alpha over g g is acceleration due to gravity so there and I'll just put it for reference here so this is when y equals to 0 aka i .e hits the ground, the ground, and then I'll just write as per my earlier video. All right, so I just put that for context. All right, so that we have this, so thus, next setup is, so thus what we notice here, If you look at this right here, so you just throw the T inside, the maximum value of T is when you just put that other side inside, because number T is between 0 and this part right here. So thus, the absolute value of X is going to be, well, less than or equal to, yeah, the absolute value of this. And the reason we're going to do this is, we're going to show that this is actually going to be within R. In other words, it's going to be, well, within this. domain basically within the parabola. So I'm going to do it right here. So this is going to be the absolute value, V0, cosine alpha. I'll put the other bracket in there. This is going to be like that. I'll put the other bracket. Like that, cosine alpha T. And then this is, and then where the T is, well, with this part here, the maximum value is going to be that. Because that's the number, that's the value when it hits the ground there. So this is going to be absolute value v0, cosine alpha, and then t is going to be 2 v0, sine alpha, over g, like that, and that's absolute value of all that. And also note now that we have is, well, we have a cosine alpha and a 2 sine alpha. This is, again, if you... have been doing a lot, seen my earlier trigonometry video, that's a trigonomic identity. That's a double angle identity. Yeah, so let's put a, like this setup here. We have, well, 2 cosine alpha, sine alpha, that equals to, well, sine 2 alpha. Yeah, so that's a sine 2 alpha. So we can replace all that. This is going to be now. the absolute value of x is going to be less than or equal to, less than or equal to, the absolute value of, well, we have a v0, that's going to be v0 squared, so there's two v0s, v0, v0, v0 squared, now we have a sine to alpha, and then divided by g, and remember the sign is just a maximum value of a sign value, is just, is one. So, this is just going to be equal to. Yeah, this is going to be, well, less than or equal to just v0, uh, uh, yeah, this is just v0 squared divided by g. In other words, this equals to just the upside value. And, uh, yeah, again, just a note right here, this right here, uh, maximum value, and this is going to be less than or equal to one, and that's because, well, if we have a, uh, just put a sign, like this for completeness, I'll just draw this right here, let's make a bubble. and draw this out. I'll just put it like this, yes. Alright, I'll just fix that up. This is going to be our sign. Yeah, it doesn't matter which I'll just put a sign out of theta right here. It doesn't matter what it is, double, whatever. It's going to be like this, and so on. That just changes the frequency of it, but the amplitude is still there. One, and then we have, this is zero. This is going to be at negative 1. Like that? Yeah, so, and again, this is, yeah, that's the maximum value is 1. Yeah, so the absolute value is gonna be, the highest one is gonna be 1, negative 1, a magnitude of that is that, and so on. Alright, so what we have here now was, well, this shows that, so this is for my Solutions Manual, writes this, instead of my usual thus, so this shows that... Yeah, so x is going to be well between, it's going to be greater than or equal to negative r, and then less than or equal to r. So, yeah. Yeah, and that's because, again, the absolute value of x is less than or equal to absolute value of r. Yeah, so then it's going to be between these, like that. So yeah, so if we look at that parabola setup there, so the x values are going to be well within a negative r and r. So now the next thing we need to show is it's going to be the y values is going to be well…