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Work the problem of projectile launched with speed vo at angle with the horizontal by using Hamilton-Jacobi methods:...

Question

Work the problem of projectile launched with speed vo at angle with the horizontal by using Hamilton-Jacobi methods:

Work the problem of projectile launched with speed vo at angle with the horizontal by using Hamilton-Jacobi methods:



Answers

Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed $v_{0}$ from height $h$

Put this question. We can draw a diagram as this is our implant plane. Okay. And this is a horizontal plane. And here is the projectile which is moving something like this and striking at this point. Okay? No, this is our horizontal line, this is our X axis and this is our Y axis. Okay, so this is X axis and this is our biases. And projectile is striking here so we can make these two points. So this will be a point and this restraints will be our cost data and these restraints will be our scientific. Okay. And the projector is launched at this the direction and it is making an angle Cheetah with the vertical and with the implant it is at 90° England. Okay. So the coordinator of the projectile is X equals U p scientist to and why equals to ut course cheetah minus gt squared by. Okay. And from the maze this amaze we can see that the coordinate X comma y are arcos cheetah minus are sinking to. Okay these are the coordinates to. What is that? Okay now substituting these equations So we get our cost data equals to this value U. T. Scientist to and minus our scientist to this will be equal to U. T. Cost you to minus Gt squared by two. So now how we have to determine this range are so we can eliminate empty. So from here time comes out to be our by you called cheetah and we can substitute this empty here. So we get minus r. Sine theta. This is equal to human plant by our by you court kita. Course cheetah minus G R E square by who you square court is. Where to? Okay so not uh We can already see that one situation is r equals to zero because artists Common. So and other situation will be GR by two. You square court is created to. This will be equals to court to course tita plus sine Theta. Okay. And this Is equal to one by scientific Okay. By rearranging day tuition. So from here range are comes out to be to us where Danny square tita 10 sq ft to develop a G. scientific. Okay, So this is the expression for the rich. Okay. So this is our required. Is that okay? So this is the distance are to the point of impact. Okay. This distance is a Okay. This is our our business. Okay.

And this problem, we're going to look at some projectile motion. Um And in particular we're going to look at an example where the information given is in terms of symbols and we're going to have to relate uh some of these symbols to what's called the range of a projectile which is defined to be the distance that a projectile would travel if it were launched from the ground, wow. And um return to the ground. The range is defined to be how far that object would travel in the X. Direction that's called the range. Um And it's given in terms of mhm. The angle initial angle of launch of the projectile and the initial speed. So we'll write that formula down, it's got some parts to it. Uh But the important thing to realize is that that range depends on not only the speed with which it was launched, but the angle was launched trump. So now to our projectile example, um what we're told is that the projectile is launched off of a cliff, mm hmm. And it actually goes twice the range in the X. Direction And it drops by one full are. And the question is um what were the initial conditions in order to make this happen? So is specifically here comes our projectile. It was launched off this cliff at some angle. Uh huh. And some speed. But what we're after is the angle that this was launched from. Um and what we know for sure is that the are in this problem is identical to uh the range that would occur um uh if that angle were used in the first picture. Um so we're going to go about this in the usual way. What we're going to do is we're going to set up our motion in the X and the Y directions. Um and we're going to have to eliminate time in those motion equations. Um And we'll get down to two equations in essentially two unknowns are in theta. And we'll be able then to solve for the theta by substituting in for our using the range formula. Okay, so motion equations uh we have delta X. So delta X in this case is equal to twice are as given in the problem and we would ordinarily write this out as that motion occurs with a constant velocity component um in the X direction times delta T. Okay. Um Meanwhile are delta Y has a very similar thing Delta Y, we're told in the problem and here I'm going to call down negative. Um So we'll say that it drops by uh huh minus are negative downwards and we'll assume that it was shot with a Y component of velocity. Um And of course there's gravity acting on it which is downwards. So just to remind her wolf will make up positive I. And to the right positive X. It's always good to set up a coordinate system. Um Okay, so what do we do with those equations? Um What we'll have to do is solve the X. Equation for delta T. Uh huh, substitute this into the Y equation. Um and also substitute in for the R. Uh huh wherever we find it. So let's take this, it's a lot of algebra. So let's just kind of take this one step at a time. First of all, we'll substitute in our delta T and see what that gives us. And then we'll substitute in for our uh kind of go through that algebra carefully. Um But what we have is basically to our over the not cosign Theta equals delta T. Yeah. And we're going to also have to square that. So let's put that in as well. Yeah. Is for R squared over V. Not squared cosine squared theta. Okay, so now we'll put that over into the um delta Y equation and that's going to look like kind of a mess at least for the time being. Um What we'll have is minus R. Is equal to and I'll leave out some of the algebra um what uh maybe not. Uh in the delta T. In the first term will cancel out and leave us with two are um sign of data over cosign Theta. And the second term is going to be the one that's going to look rather nasty. Um but the horrible cancel with the two and the denominator To give us two G. Um And then we have uh r squared that's right. Uh divided by fear not squared co site squared of data. So that that's kind of a nasty looking equation. If you were worried about things, you could check the units just to make sure that everything had units of meters on both sides. Um And I believe it does. Um And of course now we're at the point where we have to do some um messi some more messy algebra. Uh we have on the right hand side of this equation, um a non common denominator that we would have to work with. Um We could also divide out one factor of our and get rid of our negative sign. Um and we could divide by two on both sides, so a lot of playing around but we can uh at least play around with a negative sign, We can cancel one factor of our and um we can divide by two on both sides, but uh maybe more than we want to do all at once. And let me rewrite what the new equation looks like with that algebra, basically one half is equal to G. R. Over V. Not squared co sign squared of theta. Yes um minus sort of the tangent of the angle which we are going to put in just yet. Um Okay so what do we do now is um we could play around with this a little bit more um. Uh huh. So you make sure I've got this in a good way. Um Yeah we could play around with this a little bit more but I think what we are going to do at this point is sub in our range equation and to do that um An easier way to do that. We kind of see there is a product of G times are in this numerator and we can re rate our range equation with that same factor G. R. Together. Um And simply put into V not squared cosign theta science data. Um So we're going to sub in this whole expression for GR and when we do that let's see what that transformation looks like sub in for GR Yeah what will wind up with is two be not squared cosine squared data. Sorry not cosine squared cosine Theta science data. Sorry I'm already trying to cancel some things divided by you're not squared cosine squared of theater minus this little tangent which is going to kind of creep up in a couple different ways so we'll just leave that there And now for the fun we have lots of cancellation when we got rid of that a GR expression. Um And what we see is this first term looks again like twice tangent theater. It looks like basically two times sine data over cosign data. And so we wind up with a very simple expression for the angle of launch, basically one half equals two times the tension of data minus one times the tangent of data or the tangent of that launch angle equals one half. So now we have a fairly simple equation that we can solve for the launch angle simply put that in your calculator and we find out that's a to being the arc tanne of one half is a very nice angle of 26 roughly? 26.6°.. Okay. And notice that what scaled out of this was the initial speed of the launch, so it doesn't even matter. Um What uh initial speed you launched with could be 100 kilometers per hour. It could be 100 meters per second I guess. You know, you just can't break the sound barrier with that.

So they want us to find the range of this projectile with all the information they give us. Um, so that's kind of sketch the scene before we start doing anything. So they tell us where 1.5 m up off the ground, and then we're shooting some projectile at an angle, like 30 degrees with the horizon Kind of like that. What we want to do is find the range of this and remember what the range is is saying, like, how far did it travel horizontally? So what we can do for this is kind of the following. Um So if this is our position, are the horizontal or vertical position for us on? We say this is zero. Well, if the distance from the ground is 1.5, then that means down here is gonna be negative. 1.5 eso. If we set our vertical component of the position equal to negative 1.5 and then solve for T, this should give us our time for the range. And then we could just plug that back in to our position to finally get our true arrangements. Yeah, so on page 3. 15 they give us this equation right here. And so remember, this first part is the horizontal. The second is the vertical. So we're setting this equal to negative 1.5. So it was going to do that down here. So we have negative 1.5 is equal to So there's gonna plug everything in Soviet? Not is 100 of tea is just tea. Sign of 30 degrees is going to be one half and then minus one half G. I'm just gonna leave. It is g for right now. We can deal with that later. And then we have t squared. All right, Eso This is a century of quadratic, so we could just move everything over one side and then use the quadratic equation to solve. Alright, so let's do that. So I'm gonna add everything over to the remove everything over to the left side. But before I do that 100 one half, that just gives us 50. Yeah, so let's move that all over. So this is going to be now. One half GT squared minus 50 T minus 1.5 is equal to zero. And before I actually plugged this all into the quadratic equation. I'm gonna multiply everything by two. So that's going to give G T square minus 100 teeth minus three is equal to zero. Yeah, and I just want to do this toe, make the numbers a little bit prettier when we plug into the quadratic equation. So remember, the quadratic equation is supposed to be so here, this is going to say T is equal to negative B plus removes square root a B squared minus four a c all over two A. So are a is G R B is negative 100. And then our see is going to be, uh, minus three. So yeah, let's go ahead and plug all this And now. So negative b is gonna be negative. Negative. Hundreds Just be positive. 100 plus or minus the square root. So 100 squared should be 10,000 minus. So four times negative. Three. Those section would be positive. 12 and then we have times a So we have either. And then this is all over to G, right? All right. So at this point, let's go ahead and plug in what G is. Uh, so if we go back up here and look or velocities and meters per second. So that means acceleration should have meters and seconds also for its units. So that means G is going to be 9.81 m per second squared. Let's go ahead and plug this in so we have 100 plus or minus square root of All right, so it's 12 times 9.81 is going to be around 117. We had that too. 10,000. I was going to give us 10,117 0.72 all over and then two times 9.81 So 19.62 All right, so now we need to do something with that plus reminded so we could go ahead and break this up into two different ones. So it's gonna be t is equal to so 100 plus the square root of 10,117. 72 all over, 1962 and then over here this is just going to be the by not 201 100 then also over here 100. So 100 minus square root of 10,000, 117.17 all over. 19 points six to. So I want to plug everything over here on the left. I should get so exceed 10,117. Point was too big. 0.70 square root of that is around 100.5. Add that to 100 and then minus nine or divided by 9.62 on this gives me a time around 10.22 seconds. And now, if we do the same thing over here, um, this is going to give us something around. So negative point innovate develop by 90 or 19.62 So somewhere around negative zero point 29 zero here seconds. So at least from our standpoint, this time over here doesn't really make much sense. That would be like if we come back up here and look at this and we kind of fall backwards a little bit and kind of land over here, so we just kind of rule that one out on. Then we could just focus on this one. So what we're gonna do now is take this and then come up here and plug it into our horizontal. Um, So let's go ahead and do that down here. So we know that the horizontal component should be V not, uh, t co side feta. Let me just make sure I wrote that down correctly. And then we had the not equal to 100 meters per second and fate awas 30 degrees. Alright, so let's put all this in protest along with the time we just found of 10.22 seconds. So yeah, I'm just gonna plug all this in. So that means our range is going to be 100 times are times of 10.22 and then Thatta so cosine of 30 should be Route three over to and, um, if we go ahead and simplify all of this down, this should give us something around. So it's a 100 times 10.22 times Route three, and then divide all that by two. So it looks like it's around 855.0 eight and then our units with this we also velocity was meters per second. That means our distance or the range should be in meters. So this is going to be are approximate range

Here. We're gonna do some equation work. So we're told that the range of a projectile is defined as the magnitude of its horizontal displacement. In other words, the range is the distance between the launch point and the impact point. On flat ground. Because we're working with flat ground, there's a couple consequences. Let's start by breaking this down. So we're saying that the project was launched with initial speed Visa vie an initial angle above the horizontal say to I'm going to try that right here. We have a vector of the velocity. Again with magnitude Visa by the angle above the horizontal. See that if we were to break this down into its components, we have the X. Where Visa of X is equal to the magnitude Visa by times the cosine of the angle and then we'd have the Visa boy where that is equally Visa I times the sine of the angle because it's opposite the angle. Since we're working with flat ground, the Visa Y initial which is visible item signed data and will be equal in magnitude to the supply final except they'll be opposite inside initially it leaves the ground like this and when it returns to the ground this factor in this victor the exact same in X. And why? But the wise are opposite in sign. We can back that up with our third equation by saying that V sub Y final squared is equal to v. Subway initial squared Plus two times a times of change and why? Well what is the change and why between the initial and final position they're at the same height So that change and why is zero? Which means this whole term goes to zero. In other words The square of the two is equal. In other words the magnitude of these F. Is equal to the magnitude evasive way initial and final. So now that we've broken that down, let's evaluate just how much time this will be in the air. So we can say that the amount of time spent in the air is equal to the change in velocity in the Y. All over the acceleration. And we know that acceleration to be negative delta G. That Visa boy final is going to be the negative of the Visa by initial. So it's going to be negative Visa by sign. Peter minus Visa by sign. Yeah, final negative initial minus initial. All over minus G. And of course this simplifies to two Visa by sign data. All over G. This is a change in time between the leaving the ground and the impact. So now let's ask ourselves what is the range? The range since Visa Becks is unchanging. If we multiply Visa backs by the amount of time we can find the horizontal distance and we define that horizontal distance to be the range. So to get that change in X, the magnitude of it, that range will say that that is equal to Visa of x times the change in time. And now we see why we've calculated the change in time using the values that we know. Mhm. First with substitute and Visa Becks which is a Visa by times the co sign of data and now accepted to our changing time which is to Visa by times the sine of data. All over gee. Let's begin to simplify this. This comes out to Visa by squared times two sine theta Hussein. Yeah all over G. And now we're going to use a trick identity. It is a fact that sign to sign data. Co sign data is equal to the sun. Two data. Mhm. Plugging this into our equation. We can say that the range is equal to V, initial squared times the sine, two times the angle all over G. And this holds true every time that we have a projectile that starts and land with no change in the white position on level ground. Thank you.


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