Question

How do you find the maximum height of an object on it's parabolic path with three points?

Answer 1

simply :)

Answer 2

How is that?

Answer 3

all points are equi distant from the foci; we can use that, to determine a proper equation for the ellipse

Answer 4

of course, we have to consider that this forms a circle as well; which is a kind of special ellipse

Answer 5

parabolic ..... did you change that from ellipse?

Answer 6

no

Answer 7

if so, thats even simpler

Answer 8

there are a variety of methods that would work; do you have a preference?

Answer 9

no

Answer 10

hmmm, do you have 3 points to work with so that its not so absract at least?

Answer 11

yes...

Answer 12

really making me pull teeath on this one aintcha .....

Answer 13

(18,8) (45,0) (0,1.39)

Answer 14

well, the y = 0, and the x=0 points are useful for one of the simpler methods

Answer 15

ok

Answer 16

im going with an old standby; developing an equation from successives zeros

take the x values given (0,45,18) and create some zeros with them:

(0-x)(45-x)(18-x)

set up an equation such that we zero out all but one variable

y = a + bx + cx(45-x)

when x=0, this reduces to: y = a, or simply 1.39 = a

let x = 45, and y = 0

0 = 1.39 + b(45) + c45(45-45)
-1.39 = b(45) ; b = -1.39/45

this gives us so far

y = 1.39 + -1.39x/45 + cx(45-x)

so let x = 18 to solve for c

Answer 17

8 = 1.39 -1.39(18)/45 + c18(45-18)

8 - 1.39 +1.39(18)/45 = c18(45-18)

(8 - 1.39 +1.39(18)/45)/(18(45-18)) = c

therefore ....

y = 1.39 -1.39x/45 + (8 - 1.39 +1.39(18)/45)x(45-x)/(18(45-18))

is the equation :)

the wolf can simplify that as needed

http://www.wolframalpha.com/input/?i=y+%3D++1.39+-1.39x%2F45+%2B+%288+-+1.39+%2B1.39%2818%29%2F45%29x%2845-x%29%2F%2818%2845-18%29%29

Answer 18

thanks so much :)

Answer 19

there is a matrix method as well, but in the end .... we get the same results :)

good luck