1. Height of a Zero Gravity Parabolic Flight
Marissa Sittler – Period 4 – Math 1010

2. Introduction
“This lab will have you take a look at 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:

3. Writing a 3 by 3 system of equations for a, b, and c.
Using the equation provided in the packet, h=at^2 + bt +c, I created my first equation = 4a + 2b +c. For my second equation, I plugged in = a(20)^2 + b(20) +c. After multiplying everything out, I got the equation = 400a +20b + c. For the third system of equations, I plugged in = a(40)^2 + b(40) + c, and then after multiplying everything out, I got the equation = 1600a +40b +c.

4. Solving the three systems of equations
To solve a system of three equations, you must first choose two equations and choose which variable you want to eliminate. I chose to first eliminate variable ‘c’ from the equations. I chose the equation = 4a + 2b + c and the equation = 400a + 20b + c. I multiplied the first equation by a -1, so that the variable ‘c’ would cancel from both equations. After adding those two equations together, I was left with 8460 = 396a + 18b. Next, I chose the equations = 400a +20b + c and = 1600a + 40b +c. I once again multiplied the first equation by a -1 to eliminate the variable c from both equations. After adding the two equations together, I was left with 1610 = 1200a + 20b. My third step was to solve for each variable to see what they would equal. I used elimination on the equations that I had simplified earlier. I multiplied the equation 8460 = 396a + 18b by 20, and then multiplied the equation 1610 = 1200a + 20b by -18, so that the variable ‘b’ would cancel. I divided to get the variable ‘a’ alone, and found that a = =10.25.

5. Solving the three systems of equations (pt 2)
To solve for variable ‘b’, I plugged variable ‘a’ (-10.25) into an equation that I had solved for early that just had the variables ‘a’ and ‘b’ in it = 1200(-10.25) + 20b 1610 = = b 20b = b = Next, I am solving for the last variable, ‘c’ = 1600(-10.25) +40(695.5) + c = c c = 22295

6. Form a quadratic model of the data
Using the equation provided, h = at^2 + bt + c , I created the equation h = t^ t

7. Find the maximum value of the quadratic function
To find the maximum value of a function, you must put the equation in vertex form. H – = t^ t subtract over the number without a variable H – = (t^2 – ) factor out ‘a’ and complete the square H – = (t – 33.93)^2 multiply by and subtract to the left side H = (t – 33.93)^ balance the equation by adding over so that H is by itself. The max. value of a function is the vertex, so the max. value for this equation is (33.93, ).

8. Sketch the parabola

9. Reflective writing
“Did this project change the way you think about how math can be applied to the real world? Write one 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.” Response: Yes, this project did change the way that I think how math can be applied to the real world. With zero gravity ride, it is imperative that the parabola has the right curve. If it is too steep, there can be damage done to the passengers. If it is not curved enough, then you will not be able to feel the zero gravity effect.