Math Problem with a Rocket Launch

--------- #1344 - Math for a Rocket Launch?

- Attachment: photo of Kennedy Space Center Night Rocket Launch

- You give your students this photo of a rocket launch at Kennedy Space Center. The night time exposure was 2.5 minutes. How far from launch did the rocket reach its orbit at 300 miles altitude?

- The photographer was 10 kilometers from the launch with his picture perpendicular to the launch trajectory. She measured the picture frame by swinging the camera from edge to edge an getting 40 degrees.

----------- The math plan is to scale the photograph in kilometers.

------------ Measure 3 points along the parabolic trajectory.

----------- Solve the quadratic equation for the curve of the trajectory with 3 equations and 3 unknowns.

------------ With the quadratic equation calculate the distance to reach 300 mile altitude.

- To scale the photograph in kilometers we use the right triangle formed from the 10 kilometers perpendicular to half of the distance across the frame swept by the 20 degrees angle.

------------ Tangent of 20 degrees = distance / 10 kilometers.

------------- distance = 3.64 kilometers

------------ The width of the frame is therefore 7.28 kilometers. In the photo the width of the frame is 125 millimeters. Therefore the scale of the photo is:

--------------- Scale = 7,280 meters / 125 millimeters = 58 meters / millimeters

- Using this scale to measure 3 points along the trajectory where “y” is the altitude and “x” is the distance down the horizon. To simplify set the origin at the launch to get the first point of (x,y) = (0,0)

-------------- 1^st point = (0,0)

- The second point is measured at the peak of the trajectory from the photos perspective. The optical illusion is that the rocket stops gaining altitude as it curves over the horizon. In the photo we measure 40 millimeters from the launch the rocket is 40 millimeters altitude. Using our scale 2.329 kilometers down field the rocket is 2.329 kilometers altitude.

------------- 2^nd point = (2.3,2.3)

- To get the 3^rd point take half this distance and measure the altitude at 20 millimeters distance to be 26 millimeters altitude. This equates to 1.2 kilometers distance and 1.5 kilometers altitude.

-------------- 3^rd point = (1.2,1.5)

- The equation for the trajectory is approximated to be the quadratic equation:

------------------ y = ax^2 + bx + c

------------------ y is the altitude

----------------- x is the distance down field.

- Substitute the 3 points into the equation to solve for the constants “a” , “b”, “c”.

--------------- 1^st point (0,0) : 0 = a(0)^2 + b(0) + c

---------------- c = 0

--------------- 2^nd point (2.3,2.3) : 2.3 = a(2.3)^2 + b (2.3) + c

--------------- 1 = a (2.3) + b

---------------- b = 1 - (2.3) a

--------------- 3^rd point (1.2,1.5) : 1.5 = a(1.2)^2 + b(1.2) + c

--------------------------- 1.25 = a ( 1.2) + b

-------------------------- b = 1.25 - (1.2)a

- Setting the two equations for “b” equal to each other:

------------------------ 1 - (2.3)a = 1.25 - (1.2)a

----------------------- -(1.1) a = .25

---------------------- a = - 0.23

- Substituting “a” back in the 2^nd point equation:

--------------------- b = 1.25 - (1.2) (-.23)

--------------------- b = 1.53

- Rewriting the quadratic equation with the 3 known’s:

------------------ y = ax^2 + bx + c

----------------- y = -.23 x^2 + 1.53 x + 0

- To double check this result plug in the highest trajectory point (2.3, 2.3):

---------------- 2.3 = -.23 ( 2.3 )^2 + 1.53 * (2.3) + 0

--------------- 2.3 = -.23 (5.29) + 3.52

--------------- 2.3 = -1.22 + 3.52 = 2.3 CHECK!

- Now that we have the equation for the curve ---------- y = -.23 x^2 + 1.53 x + 0

We can plug in the altitude for the 300mile orbit and calculate the distance: 300 miles is 483 kilometers.

--------------- 483 = -.23 x^2 + 1.53x

--------------- 0 = -.23x^2 + 1.53x - 483

--------------- 0 = x^2 - 6.65x + 2100

- Solve this equation using the general solution for the quadratic equation;

-------------- x = ( -b + or - (b^2 - 4 a c)^½ ) / 2 a

------------- x = ( 6.65 + ( 44.2 - 8400 ) ^½) / 2

------------- x = ( 6.65 + 91.4 * (-1)^½ ) / 2

------------ x = 49 kilometers (-1)^½ , the square root of (-1) means the roots are imaginary and unequal.

------------- x = 31 miles.

------------- The photo tells us that the rocket entered the 300 mile high orbit 31 miles over the horizon.

--------------- Did anyone get this same answer? An announcement will be made shortly , stay tuned.

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