# [help] Can't figure out how to complete a parabolic equation

• Dec 5th 2012, 10:35 AM
ktelyn
[help] Can't figure out how to complete a parabolic equation
Hey everyone!

I recently did a rocket project and I have the following things calculated (attached), however I'm supposed to get my data in the format:

y - k = a(x - h)^2

I figured out the height on my own by using tangent, but I'm not sure how to get it in the correct format.

• Dec 5th 2012, 10:46 AM
skeeter
Re: [help] Can't figure out how to complete a parabolic equation
you have t = 6 seconds ... you need $\Delta x$ to get the parabola as a function of x
• Dec 5th 2012, 10:54 AM
ktelyn
Re: [help] Can't figure out how to complete a parabolic equation
Could I somehow use points to figure that out? Since the vertex would be (3,33.12) and another point I could use would be the origin at (0,0)?
• Dec 5th 2012, 02:30 PM
skeeter
Re: [help] Can't figure out how to complete a parabolic equation
the vertex is not x = 3 , it is t = 3 ...

are you wanting the parabola as height vs. displacement or height vs. time?
• Dec 5th 2012, 05:47 PM
ktelyn
Re: [help] Can't figure out how to complete a parabolic equation
I asked my professor, and he just wants the flight path (which I'm guessing just the equation?)

This is what I have but I feel like something is way off when it comes down to the equation..
• Dec 5th 2012, 05:59 PM
skeeter
Re: [help] Can't figure out how to complete a parabolic equation
how far horizontally did the rocket land from where it was launched?
• Dec 5th 2012, 06:09 PM
HallsofIvy
Re: [help] Can't figure out how to complete a parabolic equation
Another thing you can do is write your parabola in the general form $h= ax^2+ bx+ c$. You know that when x= 0, h= 0 so $0= a(0)^2+ b(0)+ c= c$, that when x= 91, h= 0 so that $0= a(91)^2+ b(91)+ c$, and that when x= 91/2= 45.5, h= 33.12 so that $33.12= a(45.5)^2+ b(45.5)+ c$ (a parabola is symmetric so we know the vertex is half way between 0 and 91 feet). That gives three linear equations to solve for a, b, and c.