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2GM So VoSuppose projectile is launched vertically from the surface r = R of the earth with initial velocity Vownere K2 = 2GM. Solve the differentialequationexplici...

Question

2GM So VoSuppose projectile is launched vertically from the surface r = R of the earth with initial velocity Vownere K2 = 2GM. Solve the differentialequationexplicitly to deduce that rft)-m ast-02GM deduce that(b) If the projectile is launched vertically with initial velocity Vo+0 >Why does it again follow that rlt)-+o as t-+0?(a) Identify the given information from the following choicesDA; Aprojectile is launched vertically from the surface r = R 2GM Initial velocity is VoDc Aprojectile is l

2GM So Vo Suppose projectile is launched vertically from the surface r = R of the earth with initial velocity Vo wnere K2 = 2GM. Solve the differential equation explicitly to deduce that rft)-m ast-0 2GM deduce that (b) If the projectile is launched vertically with initial velocity Vo +0 > Why does it again follow that rlt)-+o as t-+0? (a) Identify the given information from the following choices DA; Aprojectile is launched vertically from the surface r = R 2GM Initial velocity is Vo Dc Aprojectile is launched vertically from the surface r = 3R 2GM Initia velocity iS Vo The differential equation Solving the differential equation gives Substituting r = R at t= 0 gives r(t) = As t approaches infinity; r(t) approaches (b) Let a be the extra velocity with which the projectile will be launched, then Vo The initial velocity; Vo, is As & > 0, 2GM When the initial velocity exceeds the projectile the earth: As t approaches infinity; r(t) approaches



Answers

The range $R$ of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at $t=0$ with initial speed $v_{i}$ at an angle $\theta$ above the horizontal. (a) Find the time $t$ at which the projectile returns to its original altitude. (b) Show that the range is $R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}$ [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of $\theta$ gives the maximum range? What is this maximum 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.

All right, So we have a projectile and I'll just start with me and the information that's given to us. We know that the initial position, uh, of our projectile, I want to say r of zero. It's at the origin. Just have it at the origin. At the beginning, both X and Y equals zero, and we know that the initial velocity top place right here you not equals. It's wealth used for second in the eye direction and plus 49 49 meters per second in the direction That's the initial velocity and that. So you have both your X and your Y component right off the bat from this philosophy it's given to us. We also know other important fat, productive emotion. Your ex initial is going to be equal to the X component of velocity the entire time. That won't change, which is a convenient thing about project Ah, motion and then recording are why component of velocity. This is going to be equal to the initial velocity which we actually have conveniently right here. The initial velocity and minus we're fighting gravity with the little minus G t all right, and the X component of the position we can find that very conveniently. Um, by doing one of the first equation is you learned cinematics. X equals velocity in the X direction times time, the white component, the position of we didn't find it By using one of our very old faithful can imagine equations lossy in the Y direction times time. My guess is defying gravity, my ass. 1/2 g t sward all right. And we can find the projectiles distance Or don't distance Manitou that by using just good old with a gris If you just take your components, you square them both. You can find that radio distance over time. So now we have to do is substitute values into all these equations. And when you do that, you can get a ton of different values and you can make a table, which is what we did here. Gonna pull that out here. So take a look at that very dope, so make a table. And here you can see we have our time from 0 to 10 seconds. Velocity in the X direction. You can see states constant the whole time. Velocity in the Y direction beginning positive and then you can see it becomes negative after our reaches the max Ah heights. When you have your position over time in the X direction, which is stays positive the whole time, um, increasing from zero. And then for the y direction or the Y position, you see, it increased from zero gets to its max, heightened and decreases. And in the radio position, you can see conveniently matches. Um, the X motion. For the most part here you see crops all the way across and ends at the 1 20 as well. And that's just a vector that is essentially tracking whole motion. So it's not gonna be the same as the X, but it's gonna have that same starting value and ending value as the X that you can see there with zero on the top of the 1 20 on the bottom. All right, now we want to also consider something here. We want to consider the fact that the velocity is not perpendicular to the displacement where the distances Maxwell. So there's only gonna be some component. The velocity along are opposite to the direction of this radio factor. This component of velocity will either decrease or increase the value off our magnitude of our radio actor. It cannot be maximum. So this means the velocity back. You must be perpendicular to our that actually don't like this. The velocity vector has to be perpendicular to our that's perpendicular to our, um, when the magnitude of the radio component is maximum is Max important, important fact everything to work with. All right, so now at this point, we're going to starting with what we're given here. At the beginning, we had our initial velocity and we had our acceleration, which we know is right. My acceleration here cell ratio equals minus nine point each one meters per second squared. And we want to consider the magnitude of displacement vector. We maximum at time t. So let's I use this year to describe Sparks is out of here, and we'll do velocity equals V nuts plus acceleration, times, time and we can plug in what we know for this. We already know what the initial velocity is, and we know what the acceleration is. We just have to make sure we put a minus in for the acceleration, and then we're gonna be able to come have just time left over So we will get G equals plugging in Vina here this factor first and then plugging in the acceleration second and cleaning it up. That's why we get 12.0, in the direction plus using my Prentice easier. Right? But we have 49 0.0, in the J direction minus the 9.1 meters per second square times t also in the direction. So that's why we have these guys couldn't hear together And we can also get a displacement vector are doing a peanut T plus 1/2 eighties word. And when you plug, you know all the stuff that we have that we have the initial velocity vector right here and we also know that the acceleration is the negative 0.4 meters per second squared. So when you plug all that in and you go ahead and clean this up, you look at something of you sure are equals. Well, you know, t I had waas 49. Well, which is that j 60 minus the four point on which is part of 1/2 of the acceleration 4.9 xi squared. Well, this isn't a J direction now maximum displacement. The velocity in the position vectors are perpendicular. Hence, the doc product will be zero very important fact here, the the or at maximum displacement, the velocity and position vectors here perpendicular. So these guys zero doc wallet. So when I move this over here in the game, this up here is not years 12 I waas or 49 minus 9.1 T. And, of course, we have his full thing with just boxes in right here. It started dark products with our radio vector, which we found here really running out of space there. And this is, of course, in the J direction. This last guy. And then when you run through this Doc Products, um, you run through it, find that equals zero seconds, 5.69 seconds and 9.3 seconds. Lovely around the displacement at zero seconds in 9.3 seconds, um, are actually the minimum. And so the maximum displacement occurs. Actually, just had this middle one here is 5.69 seconds. All right, so we want to substitute this value of tea into the equation of displacement, which is this equation right here. And what I'm gonna do much. You gonna take my, uh, my lowly table and witness to the sides? We can actually have soul more space here, you know, right there. All right, so we're gonna take our our equation here, or are, And we can now plug in the 5.69 seconds and each one. He's my direction. Plus the 49 you know, times 5.69 minus four point times 4.69 squared. All right. And then when you do this, you don't figure out both components, which is 68.3 and direction and 120.2 in the J direction. If you want to find the magnitude, we will use the same thing over here. What we've that before. So the magnitude of the radio vector being 68.3, I swear, plus 320.2 squared. And then you want to get roots of this, and you're gonna gets roughly 100 on 38 years now just to kind of build long. We just did here. I had to use the fact that for maximum displacement, velocity, vector and the displacement vectors of perpendicular. That's always what we had that dot product that fee dot are down here is we have to use that fact about the fourth organ alley essentially with other than the perpendicular at those specific points now of two vectors of perpendicular. You know that that part of zero So first I had to consider that time t the displacements, baby Maxwell. Um And then I calculate the displacement in the velocity of tea. And then I took stock product and said the dot party with zero since they are gonna be perpendicular and will this solve for time and in knowing time you're employed into our radio expression and, um, find both components you x and the Y components, and then go and find the magnitude here, which is what we did.

So let's start answering any concept questions 1st. So for a question a it asked if the horizontal component of the initial velocity just being at X is proportional to the initial velocity. So we know the formula for horizontal component of initial velocity? Yes initial velocity multiplied by call sign of the initial. So as you can see from that equation then we can say that the horizontal component of initial velocity V. X. Is directly proportional to Vienna which is the initial velocity magnitude of initial velocity. Now question be asked if the time of like is directly proportional it's proportional to initial velocity. The formula for time like is this twice the initial velocity multiplied by sign of the angle over the acceleration due to gravity. As you can see from the formula we can say that the time of flight is directly proportionately initial philosophy. No questions see us if the horizontal range is proportional to the initial velocity R squared of the initial velocity given that Okay, it was stated That the ranges equals two. The product of the horizontal component of initial velocity in the kind of life. So from our equation from party Indy then we have Vienna causing data multiplied by to Vienna scientist and not over G. So simplifying then we have the nut squared two because I wanted to not scientists and not over G. And from the economic identity to casa and data. Mhm. Science data is equals to sine two. Theta. So our equation becomes like this. So as you can see the horizontal range is directly proportional to spread of the initial velocity. Now for the problem question. Mhm. It asked what if the range 623 m and now the initial velocity is doubled. So we have new initial velocity but the angle is the same. So taking up in that's not going to take a nap. What is the new range? So again, if R. S. Equals to Vienna Spread signed to take a nut over gee in his Echostar 23 m then are some any Sequels to? We're not and squared sine to beat up not in over G. So inserting the value of words we know, then we have to Vienna quantities grade modified by sign to data nuts. Is the under are the same over G. So simplifying. Mhm. Then we have four Vinod squared sine two Theta nut over G. But we know that This quantity here is just 23 m Equal to 20 of four times 23 m is equals to 92 m. As you can see, The new range is 92 m.

Okay, so this is just simple third record question where we have to find first the amount of time it takes these, this ball, this projectile to go from its initial spot to its end spot where the same altitude we have the constant acceleration of gravity. And it's negative. 9.12 meters per second squared. Are we replace that 181 g. So just negative, g really? And then we have that. So I want to find out the amount of time Let's split this sort of path into two sections. We have going up and we have going down. Okay, let's do going down blue now. The thing to know is that going up the green path and going down the blue path actually takes exact same line of time. So all we need to do is calculate degree path and multiplied by two. So how long is the green path gonna take going up to the peak? Well, we know that at the peak, the vertical velocity is going to be zero. That's just what happens at the peak, right? So let's use this for a V y equals two V in initial. Why plus vertical acceleration times time. So let's deconstruct us. We know that V buys gonna be zero. We know that the i y using you know, the strangle is going to be able to v i science data and then we have negative g as the vertical acceleration and we have time. So let's all for time. We'll have time is equals to if we sort of rearranges equation is gonna be weise. I signed data over Jean, and now since we're absolutely multiplying by two, we have that the total time it's just gonna be equal to to v I Science data over Ji. So that's your answer party. If a part b, we have to find the range. Okay, so we have a vertical horizontal velocity v I X. That's gonna be the same throat, right? So Vieques is always going to be equal to the initial horizontal velocity, and that's gonna be equal to V I Coast State. All right, we can just use the stronger here to figure that out. Now that we have this, we also have the time. And since for a constant velocity, the distance is just gonna be good to the velocity times of time. And so the range is just gonna be able to v I coast data times a time which is equal to, you know, over here is gonna be to V. I signed data over G. And now that's that's ah, you know, rearing. Some things will, but the two sides Data Coast data at the front Well combined the Steuby eyes together to get V, I squared and we'll divide the politic buggy. Now there is ah, important to economic signal metric identity that were given were to sign or sorry yet to science data. Coastline data is also gonna be equal to signed tooth data. So if you replace this with this, get r equals to sign tooth data. The I squared over g and there's ring. Okay. And lastly, we have to figure out what value of data is gonna give us. Uh, the greatest possible rich. Okay, so let's examine this form of forage. Okay? If we have this formula, what value of data is going to give us? Ah, the greatest value of signed tooth data, while the maximum value that the sine function can ever give us Okay, the maximum value could ever give us is one. So what value off there is going to be equal to one? Well, we know that sign. 90 degrees is ableto one. And so if we put data as 45 degrees, so then becomes signed two times 45 degrees, which is 90 will get What? So the best possible angle for the maximum range is going to be 45 degrees. And what is going to be that range? Wallet just replaced scientific data with one to get our equals. Two v. I squared over G. So there's your max range. All that are Max and yeah, that's your angle, That's it.


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