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Height of a Zero Gravity Parabolic Flight Math 1010 Intermediate Algebra Project #2 By Kelsey Evans P.2

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Have you ever wondered what it might feel like to float weightless in space? One way to try it out is to fly on a special aircraft that astronauts use to train for their trips to space. Both NASA and Russian Space Agency have been flying these for years. The way this is accomplished is to fly to a high altitude, drop down to gain speed, and then start a large parabolic path up in the sky. For a time ranging 10 to 20 seconds, along the top part of the parabolic flight, an environment simulating zero gravity in created within the plane. This effect can cause some nausea in the participants, giving rise to the name Vomit Comet, the plane used by NASA for zero-G parabolic training flights. Currently there is a private company that will sell you a zero-G ride, though it is a bit expensive. This lab will have you take the parabolic path to try to determine the maximum altitude the plane reaches. First, you will work with data given about the parabola to come up with a quadratic model for the flight. Then you will work to find the maximum value of the model. Now for the data: Height of a Zero-G Flight t Seconds After Starting a Parabolic Flight Path Time t in seconds Height h in feet

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To find the quadratic model, you will be plugging the data into the model h=at^2+bt+c. The data points given are just like x and y values, where the x value is the time in seconds and the y value is the altitude in feet. Plug these into the model and you will get equations with a, b, and c. Part 1: write your 3 systems of equations for a, b, and c. We will plug the data points into the equation: h=at^2+bt+c 23645= a(2)^2 + b(2) +c You plug the first set of data into x and y for y and 2 for x =a(20)^2+b(20)+c You plug the second set of data into x and y for y and 20 for x =a(40)^2+b(40)+c You plug the third set of data into x and y for y and 40 for x.

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Part 2: solve this system. Make sure to show your work. Step 1: Take two out if the three equations from part =a(2)^2+b(2)+c Take the first to equations =a(20)^2+b(20)+c 23645=4a+2b+c Simplify 32015=400a+20b+c -1(23645=4a+2b+c) Take the Frist equation an times the whole thing by =400a+20b+c We do this so that the two cs will cancel out =-4a-2b-c The two cs will cancel out 32015=400a+20b+c 8370=396a+18b This is what we are left with.

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Step 2: Take the last two equations from part 1 and set them up and solve them the same way as step =a(20)^2+b(20)+c Last two equations from part =a(40)^2+b(40)+c 32015=400a+20b+c simplify 33715=1600a+40b+c -1(32015=400a+20b+c) Times the first equation by -1 so that the cs will cancel out =1600a+40b+c =-400a-20b-c The cs cancel out =1600a+40b+c 1700=1200a+20b You are left with this equation.

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Step 3: Take the two equations that we solved for in steps 1 and =396a+18b Find something that either the a or the b will be able to make the same on both equations. 1700=1200a ( 8370=396a+18b) The bs will be able to cancel each other out so you find a common multiple and -9(1700=1200a+20) then you multiply both of the bs by a number to make it the common multiple. - in this case you would times the top one by 10 and the bottom on by =3960a+180b simplify = b the bs will cancel out =2880a This is what we are left with but it can be simplified down even more. You divide both sides by a=23.75 This is out solution for a

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Step 4: use the equation that we got from step =1200a+20b plug the solution for a into the a in this equation. 1700=1200(23.75)+20b Simplify 1700= b now solve for b - subtract 28500from both sides =20b Divide both sides by b=-1340 This is what b is equal to.

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Step 5: Take the first equation from part = a(2)^2+b(2)+c now plug the answers for a and b from steps 3 and 4 into this equation =23.75(2)^2-1340(2)+c Simplify 23645=23.75(4)-1340(2)+c 23645= c 23645=-2585+c add 2585 to both sides c=26230 Now a=23.75 b= c=26230

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Part 3: Use your solutions to the system from part 2 to from your quadratic model for the data. Plug the answers for a, b, and c into this equation: Y=ax^2 +bx +c You should get this: Y=23.75x^ x

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Part 4: Find the maximum value of the quadratic function. Make sure to show your work. To find the maximum value you have to use the equation from part 3 and put it into vertex form. The equation for vertex form is f(x)=a(x-h)^2+K we Get to this equation by completing the square. Y=23.75x^2-1340x now subtract from both sides Y-26230=23.75x^2-1340x now take out from the right side Y-26230=23.75(x^ x_______) now divide by 2 and then square the answer. After you do that put that number into the blank space. Y-26230=23.75(x^ x ) what ever you do to one side you have to do to the other so you multiply by and then add that to the left side. Y =23.75(x^ x ) Then you complete the square and move back to the right side. Y=23.75(x-28.21)^ this is the equation in vertex form. The max value is the number that is represented by k In the equation above. Maximum value is

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Part5: Sketch the parabola. Label the given data plus the maximum point. A good way to start labeling your axes is to have the lower left point be (0,20000) o This is the maximum point o (2,23645) (20, 32015) o (40, 33715)

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Part 6: Reflective writing. Did this project change the way you think about math and how math can be applied to the real world? Write a paragraph stating what ideas changed and why. If this project did not change the way you think, write how this project gave further evidence to support your existing opinion about applying math. Be specific. I think that this project was a good way to even further my opinion that math is used in the real world. This math project showed that you use math to solve physics problems. This math project showed that math is used to determine how high an airplane must go in order to achieve zero gravity and how long that zero gravity would last. I thought this project was a very interesting way to apply what we have been learning into a problem that people are solving everyday in the real world.

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