repeat-problem-26-under-the-assumption-that-the-rocket-is-launched-from-a-platform-five-meters-above

Name: repeat problem 26 under the assumption that the rocket is launched from a platform five meters above
Uploaded: 2020-06-16
Duration: 7 min 25 s
Channel: Numerade
Description: Repeat Problem 26 under the assumption that the rocket is launched from a platform five meters above the ground.

Question

Repeat Problem 26 under the assumption that the rocket is launched from a platform five meters above the ground.

Michael Jacobsen · Accepted Answer

here, we&#x27;re going to be dealing with a rocket that needs to reach a maximum height of 90 meters. We need to determine what the initial velocity ought to be. Have a Maya time. It&#x27;s going to take to reach those 90 meters. We&#x27;re further going to suppose that the rocket launches from a structure that is five meters high. Let&#x27;s start with the differential equation that governs its motion. That would be due to why d t two is equal to G or the second derivative of why is equal to G, which is acceleration due to gravity. Also, since a rocket launches from a structure five meters high, we can say that why it zero or the height at time zero must be five meters on our gold, then is to say, what should the initial velocity be? In other words, what is D Y t. T equal to next will solve the differential equation. By taking anti derivatives, we&#x27;ll get that d Y. D t on the first step is equal to g times T plus a constant C. We don&#x27;t have any information on the first derivative for initial conditions, so we&#x27;ll just proceed to the very next step of taking the anti derivative of what we&#x27;ve just produced. We&#x27;ll obtain that Y is now equal to G divide by two times T squared plus C times t plus a constant integration deep and that gets us set up. So far, the next thing to do is solve for either one of these constants. By the given initial condition that why it zero is equal to five meters we should be able to solve for D. So let&#x27;s write this out. So why it? Zero by substitution substitutes in the quantity zero in place of tea so we&#x27;ll obtain that G over to time. Zero squared plus C time zero plus de is the left hand side, Wyatt zero. But why it? Zero must be five meters, so five is going to be equal to D. And now the equation that governs the motion of the rocket is like walls. G divide by two times T squared plus C times T plus five. For the next step, recall that the rocket will reach its maximum height when velocity is zero. In other words, the max height is obtained when de y duty become zero, but our expression for D. Y. T t is given here, so D Y t T equals zero now means that zero is going to be equal to G Times T plus C. And if we saw for the variable T, then T will be equal to negative C divide by G. So now we know the time at which the maximum height is obtained in terms of the variable. See, we&#x27;ll be able to use that information to solve for C. But C itself is the initial velocity that we would like to find. So for the next phase of this approach, we go back to this equation that governs the motion of the rocket. And we can right now why, at a time of negative see over G will be equal to through substitution. G divide by two times negative C over G squared plus see times time, which is also negative. See divide by G, then finally plus five, then the trick to this step is negative. See, over G was the time at which max height was reached, but the Max High is also stated to be 90 meters. That means the left hand side of this equation can be written in with just a 90. The next thing we want to do is simplify the right hand side and then solve for the constant C. If we work out, the very first term will obtain C squared in the numerator divide by two G and the denominator plus negative C squared divide by G plus five. So let&#x27;s do a couple things at this step. Let&#x27;s subtract five from both sides of the equation, so we&#x27;ll have a 85 equals down the right hand side of this equation. Let&#x27;s insert two here as well as here, so that we have a common denominator. Our con women denominators to G. We&#x27;ll have C squared minus two c squared in the numerator, so just a negative C squared. Overall, if we next multiply both sides of the equation by two G, we&#x27;ll obtain that 170 times G is negative. C squared. Multiply both sides by a negative one and now negative 170 g is equal to C squared. Now we can solve by taking the square root of both sides. The solution for C is that C is equal to the square root of naked negative 170 times G. But since jeez acceleration due to gravity, we also know that G itself is approximately equal to the quantity negative 9.807 meters per second squared. So negative 9.807 And inserting all this into a calculator tells us that see is approximately equal to 40.83 meters per second. So that tells us next that the initial velocity is de Y t t. At zero. If we insert zero into this expression for velocity, we just get G time zero plus c, and we just found that C is equal to 40 point 83 meters per second. So if we set our initial velocity to that quantity will reach the height of 90 meters. And the problem also wants to know how long it&#x27;s going to reach that maximum height. To find that require time, we can go back to this expression here and say that the time required is t equals negative. 40.83 Divide by G, which is negative 9.807 and this becomes four point 16 seconds

Hayden Woerner · Answer

Okay, so as we weren&#x27;t problem 73 Umm, we are looking at Anak cell oration. Um, so it says after five seconds, the rocket reached a maximum height of 412 feet. Find the initial velocity and the height of the rocket. So we&#x27;re looking for we need the velocity function and we need the position function in order to do that. So what? We know if there&#x27;s a maximum AT T equals five, we know that at that point the velocity is equaling zero. And we also know that the position is 412 right there at that same value of teeth. So we know that a of T for, um, a falling object is negative 32 feet per second. So if we want to find the velocity function, we can just integrate that with respect to time. And that&#x27;s going to give us negative 32 t plus C. And so we need to find, See? And what we know is we know that the velocity equals zero when t equals five. So 32 times five. Let&#x27;s see, that&#x27;s 1 50 plus tens or 1 60 So, um, zero is equal to negative 1 60 plus c soc is equal to 1 60 So our entire velocity function is negative 30 two t plus 160. So if we want to find the initial velocity, I&#x27;ll do that. Over here on the side, we plug zero in for tea. So the initial velocity is 160 and our units is feet per second. All right, so now I&#x27;m gonna continue back and deal with the position function. So if we want to find S a T, we need to integrate V A T with respect to time. So I&#x27;m gonna integrate this entire function. So s of tea is going to be when we add one to the power and divide by that power, we&#x27;re gonna get negative 16 t squared plus 1 60 t plus C. And it tells us that the uh rocket reaches a maximum height of 412 at time. Five. So five squared is 25 1 60 times five plus c. And so if you continue to solve that equation until you get sea by itself, um, you will find this value of C to be 12. So our entire position function is negative. 16 t squared plus 1 60 T plus 12. Which means that our initial condition are our initial high s of zero is going to be 12 feet.

Sajin Shajee · Answer

okay. Raising a height for age of tea which is equal to minus 16. Where was the zero g? Must each zero first were given each of five keys Agent by prime region equals here the visits in maximum along with also each of five Equalling 412. So H Prime team go ministry too Waas, you hear h prime of five people to minus 1 60 plus easier with just the clothes here these years equals 1 60 No, we&#x27;re gonna sell for the to using this the busier here age of key Each of five is gonna be equal to minus 400 b zero times five plus each steer which is people to 412. If the zero is on 60 then I&#x27;ll be 400 plus 800 plus a nine queen 412 eight zeros equals 12 So then wait Simply become this is all enough heat The creation of age he cooling man 16 he squared plus 160 e plus 12 Sears 12 ft can finish velocity V zero is 100 16 ft per second

Angela Guo · Answer

this question is a straight plug in using the formula we found in the previous problem. So I have written the results here, and I have also written down the constant. Okay, so for this question were given is that, um are is equal to 20 kilometers. But before you actually just plug that straight and make sure you cover the two meters first, so our is actually going to be 20,000. So now we have all the numbers that we need to put into this equation. To sulfur be not so. Once you do that, you&#x27;re going to get to be not is going to be equal to 626 meters per second.

Donald Albin · Answer

a rocket is launched vertically upward from Earth&#x27;s surface at 5.1 kilometers per second. I wrote that down, and I also converted it into meters per second and were asked for its altitude. This is a good question to use, um, conservation of energy. So the kinetic energy, plus the potential energy initial equals the kinetic energy, plus the potential energy final. Well, we have set, um, potential energy to be zero at infinity. So the potential energy initial and potential energy final are both going to have a value. Neither of them is going to be zero. Um, our initial kinetic energy has to do with the launch velocity, But our final kinetic energy, thank goodness is zero. So we don&#x27;t have to worry about that turn. So, um, are kinetic energy initial weaken right down as 1/2 I m v squared. All right. Our potential energy initial potential energy is negative. G i m small m over our so I&#x27;m gonna write minus G. I m. So this is, uh, launch from the Earth&#x27;s. That would be the mass of the earth. M is the, um, mess of the rocket. Interesting Hope that cancels out but over our So this is on the earth. So that would be the radius of the earth. OK, negative. GMM over our okay equals negative. G mass of the Earth. Small am again, which is the mass of the rocket over our But this time it&#x27;s are of the earth, plus the altitude. I&#x27;m going to write a l t for altitude. No, I&#x27;m not. I&#x27;m going to write a check for altitude. Let me go back. Each h is going to be the altitude. So if you think about it, it starts off one the radius of the earth And then it goes even farther away by a distance h which we would call the altitude. Now I notice that the mass of the rocket massive the rocket massive the rocket that completely cancels out. So the mass of the rocket is negligible. And that&#x27;s a good thing, cause we&#x27;re not given that in the question. Um, so we&#x27;re going to figure out the altitude, Uh, so trying to think about whether I wanna solve this whole thing to make it say h equals or if I just want to put some numbers in there and then evaluate it. Um it might look a little funky to make it say h equals, but guess I&#x27;m gonna do it anyway, So we&#x27;re gonna multiply by one over negative g m on both sides of the equation. No, I&#x27;ve run out of room to write that over there. So I&#x27;m just going to write negative one over GM on both sides of the equation. That&#x27;s going to cancel out the negative one over GM. And now I&#x27;m going to write the right side on the left. So that&#x27;s gonna leave me with changed the blue here, one over our sub e mm plus H equals. Okay, so our first term is now going to be negative. We still have a V squared, but negative one over GM times 1/2 is, um, negative, which I have the negative there already to gm. Okay, So negative one over GM times. 1/2 B squared is V squared over to GM, and it&#x27;s negative. All right. Negative times. Negative is a positive. And so GM, divided by GM, cancels out. And I&#x27;m just left with one over the radius of the earth. Okay, Now we have to invert that, and then we have to subtract the radius of the earth from both sides. So I&#x27;m gonna go up the top here and erase. I could and probably should get a common denominator. But I&#x27;m not gonna do that. I&#x27;m just going to write Whoops. I&#x27;m just going to write one over. Ah, intended that to be blue. Try that again. I&#x27;m just going to write one over and then I am going to write. Going to be able to see that&#x27;s a one. Invert the right side that&#x27;s going to give me and I&#x27;m gonna flip it around. One over r E minus V squared over to G I m. Okay, so all I did was I inverted the right side, and now I&#x27;m going to invert the left side, which is going to give me are sub e plus h. And so if I would subtract RCB from both sides then it would tell me h equals that Ah, rational, complex fraction there minus arson E. So I&#x27;m gonna switch this around a little bit. Minus are sub e equals h. All right now I gotta put numbers in for that. Gonna have to create some rooms. Ah, go ahead and erase this. Don&#x27;t know if I&#x27;m gonna have enough room, but give it a try here. Okay, One over one. Over. Radius of the earth. 6.37 times, 10 to the six power minus V square. That would be 5100 squares. Is V squared over to G 6.67 times 10 to the negative. 11th power M Capital M. And we&#x27;re still talking about the earth. So the mass of the earth, 5.97 times, 10 to the 20 fourth power. Okay, one over all of that. And then after I&#x27;m done and I don&#x27;t have any room, but after I&#x27;m done, and I&#x27;m going to write this up here and they have to subtract the radius of the earth. All right, let&#x27;s go ahead and put that all of that into a calculator, and that should give us the altitude. All right. I did all of that. And then I subtracted like I said, what I wrote in red there are e RCB is 6.37 times 10 to the six power. Um, I don&#x27;t have anywhere to write this, So just gonna clear some space here. That gave me an answer which was h. Try this again. Each equals you go back to my calculator so I can see it. 1.6 seven times 10 to the sixth power kilometers. That&#x27;s about the order of magnitude that I expected. It is less, well, 1/3 of the radius of the earth a little bit less than that. So I think this answer is reasonable and makes

Paul A. · Answer

So, uh, once again, we have an extension off the previous problem. And Aziz, you can recall in the previous problem we had a rocket that&#x27;s launching off with an initial velocity. And and of course, when you think about the initial velocity, that&#x27;s the velocity due to the exhaust. So these are the exhaust velocities and this was 1500 meters per second and it takes off. Had an initial mass, we computed that to be approximately 809 kg. You know, that was the initial mass on. Then the rocket takes off and goes all the way up and the travel time from takeoff up until, ah, specific point in space is delta T equals 2 10 seconds. So after 10 seconds, theme. The rocket has specific masters, so the rocket has lost. Um, it&#x27;s original mass from the exhausts, so the the exhaust it&#x27;s losing its mass. And so the difference between the initial mosques and the current masses 100 kg, as you can recall, and also another piece of information that we&#x27;re given is the fact that it reaches a speed off 100 m for second. The initial velocity is 1500 m per second. And now in this new problem is, well, kind of like in a space station. We&#x27;re in a space station. And the, um uh the gravity is 0 m per second squared, recalling the other case. The acceleration due to gravity was like 9.8. But in this case, you know, we the rocket is taking off from a space station, but the parameters are still the same. S o. You know, the only difference is that we have no gravity in the space station. So gold in this particular problem is way. Want to figure out what the initial mass is gonna be? So what&#x27;s what&#x27;s am I which is the initial Hamas? If the rocket Uh huh takes off from space station, staking off from the space station&#x27;s of wanna look at the initial mosque again? You know, we&#x27;re still going to go back and use the previous formula that we have. So the change in velocity is you, Alan, off M I. Which is the initial mass over the current moss. But if you can recall from the previous problems, something unusual happened. Well, I mean, I don&#x27;t want to say it&#x27;s unusual, but it&#x27;s It&#x27;s just part of the problem. Something happened in the previous problem, and it&#x27;s the fact that we had minus G. Delta T. But in this instance, in this new problem, we don&#x27;t have this part because the we&#x27;re dealing with the space station and so the acceleration due to gravity or these negligible gravity in the space station. So that&#x27;s what we&#x27;re having in this new problem. And, um, that reduces the equation that we&#x27;re dealing with you is the exhaust exhaust speed That&#x27;s you, uh, m I is the initial mass. Am I? The initial mass m is the current moss, uh, or current instantaneous must, if you want to see that instantaneous, instantaneous, instantaneous us. Um, And if we if we bomb the fuel, uh, we know that the instantaneous mass M is gonna initial is gonna call the initial mask under the mask from the fuel. So this is this is the fuel that&#x27;s being bond. That&#x27;s that&#x27;s the mosque right there. And so Delta a V is you, Ellen, off the initial HMAS and also the instantaneous months, which we were right in the new former and so one assault fall. The initial mass we divide both sides by you. So, Ellen, off the initial months off initial mass minus the mass of the fuel that&#x27;s lost equals two Delta V over you. We can eat both sides, and that&#x27;s going to cancel the Ellen. And so we have the initial months off. The initial mass minus final months equals to eat out of e all over you. We want to multiply both sides by this value, so they instantaneous mosque. This is the instantaneous must. So the instantaneous masses multiplied on both sides and we get a formula for the, um for the initial months. But then we can use distributive property to isolate the initial masses, as you can see, isolating the initial masses because we want to bring it to the other side. So when we subtract, um, when way can add he Delta u final mass of the fuel, we are it on both sides. So this is just pure algebra. Are these to cancel? Then we subtract the initial mass from both sides such that these to cancel on. And that simply means that we have the e. Delta V overview. Initial mass minus initial mass equals to eat Delta the of the U final mass. That&#x27;s what we have right there Can pull out the initial months. So we have it out of the over. You, like this one equals two e Delta V over you, the mass of the fuel. Then we divide both sides by this&#x27;ll value the isolate the initial mass. So the initial mass becomes, um, the initial moss becomes the months off the fuel must off the fuel times E Delta V off you all of e delta view of you minus one. That&#x27;s what you&#x27;re seeing right there. So just to make sure we have an understanding off the problem, we had to do e on both sides. We had t e both sides on the moment. We did e both sides, uh, these two canceled out, so we had I m I m I minus MF the Ellen&#x27;s. They canceled out. So we had, um we wanted to reorganize parts of the problem. And so at this point, we have the fuel mass off the fuel, which is 100 I can recall on. The problem is, say the fuel is 100 so we have 100 times e delta v is the change in velocity That&#x27;s 100. Meet us for second and then we divide that by 1500 which is the initial velocity. And so on this side we have the same thing. The 100 m per second over 1500. I need a second and then we want to subtract one from that. So the last step of the problem is just the simplification process where we way can have e. That&#x27;s when you know. Uh huh. We have erased, too 100 to fight by 1600 that then we want to divide that by he raised to 100. You know, when these wants is just to make sure that we&#x27;re doing the right thing. Some 100 divided by 1500 on minus one, Mhm. Then all of that will multiplied by 100. So we end up getting approximately, uh, you know, 1500 0.5555 And when you simplify that, you get the initial mass to be exactly the same as 1551 kilograms. 1551 kilograms. This was also kilograms. So again, you know we had a problem. Was similar to the past problem. And we had the rocket. There was no gravity. So that means in the equation for simplifying for the initial mass. We got rid off this portion of the equation and then used algebra, uh, in both sides so that we can cancel the L N and then reorganize the equation toe, isolate three initial mass and then plugging in the specific values to get the mass to be 1551 kg. I hope you enjoy the problem. Feel free to send any questions or comments and of the wonderful a wonderful day.

Problem

An object that is initially thrown vertically upw…

Problem 27 Hard Difficulty

Repeat Problem 26 under the assumption that the rocket is launched from a platform five meters above the ground.

Answer

$=40.83 \mathrm{~m} / \mathrm{s}$

Related Courses

Differential Equations and Linear Algebra

Related Topics

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Kayleah T.

Samuel H.

Michael J.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra

Catherine R.

Kayleah T.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$=40.83 \mathrm{~m} / \mathrm{s}$

Differential Equations and Linear Algebra

Catherine R.

Kayleah T.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

Catherine R.

Kayleah T.

Samuel H.

Michael J.

Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra