What gives?

projectile launched @ 10.0 m/s @ 45.0 degrees

figured everyhting out go these three points
I'm trying to apply my Algebra 2 class stuff to it to get the equation of the perabala

(0,0) (5.1 , 20.4) (10.2 , 0 )
then i use matrix to find the values and my equation i get is

y = 15.2X^2 + 276X + 797

What gives?

It is easier to write down x and y as a function of time and eliminate the time variable, you should obtain:

y = -g/v0^2 x^2 + x

where v0 = 10 m/s

If you have three points that are on a parabola and you want to solve for the equation of the parabola, then you should not write down three equations for three unknown parameters. That's a very inefficient way to find the parameters.

You can write down the equation immediately, without solving any equations at all! Suppose that the three points are (x1, y1), (x2, y2), and (x3, y3). Then, we want to somehow write down a quadratic formula in which we can just edit in the values x1, y1, x2,...etc, and it will be the correct formula. But this means that if we change, say, y1, and leave all the other parameters the same, it will be the correct formula for the new situation in which only y1 has changed.

So, the structure of the formula must be such that changing y1 will not affect y(x2) and y(x3). A quadratic formula that is at x2 and x3 is, of course, given by:

(x-x2)(x-x3)

This term should control the value of y at x1, so we multiply it by

y1/[(x1-x2)(x-x3)]

To get the values at x2 and x3 correct, you can proceed in the same way. The general formula is:

y = y1(x-x2)(x-x3)/[(x1-x2)(x1-x3)] +
y2(x-x1)(x-x3)/[(x2-x1)(x2-x3)]+
y3(x-x1)(x-x2)/[(x3-x1)(x3-x2)]

Let (x = 0, y=0) be the starting point at t=0.

I do not know what your
(0,0) (5.1 , 20.4) (10.2 , 0 ) represents, nor what a matrix has to do with it.

x = 10 cos 45 t = 7.07 t

y = 10 cos 45 t - (1/2) g t^2
= 7.07 t - 4.9 t^2

= x - 4.9 (x/7.07)^2 = x - .098 x^2

That is the parabola equation.

y = 0 at x=0 and t = 1.44 s, when x = 10.2 m

ok i'm lost heres what i got for all of my values

Vi = 10.0 m/s
angle = 45.0 degrees
Vyi = 7.07 m/s
Vxi = 7.07 m/s
g = -9.81 m/s
Vyf (@ top) = 0
delta t (1/2) = .721s
delta t = 1.44s
delta y = 20.4 m
delta x = 10.2m
delta x (1/2) = 5.1m

that's how i got my points
delta X (1/2) and delta y
giving me the vertex
(5.1 and 20.4 m)
(0 and 0) being the start
and
(10.2 and 0) for when it reaches the bottom

my work

tell me what i did wrong
(sine < = opp/hyp)(hyp)
hyp sin < = opp
plugeed in numbers
10.0 m/s sin 45.0 degrees and got 7.07 m/s

sense the angle is 45 i can conclude the the Xvi is the same right? I did the work anyways

(cos < = adj/hyp)hyp
hyp cos < = adj
10.0 m/s cos 45.0 degrees
Vxi = 7.07 m/s

Vyf = Vyi + g delta T
0 = 7.07 m/s + (-9.81 m/s^2) delta t
subtraced the 7.07 m/s from both sides

- 7.07 m/s = -9.81 m/s^2 delta t
multiplied out by 1/-9.81 m/s^2

-7.07 m/s
_________
-9.81 m/s^2

and got delta T half = .721 s
multiplied by two
got total delta 2 of 1.44s

then i used this equation
delta y = Vyi delta t - 1/2g delta t^2
pluged in numbers
delta y = 7.07 m/s (1.44s) - 1/2(-9.81 m/s^2)(1.44s)^2
and got for an answer
dleta y = 20.4 m

then i used this equation
delta X = Vix delta T
7.07 m/s (1.44s)
delta X = 10.2 m
divided that in half for midpoint
delta X (1/2) = 5.1m

then already stated i got the three points

(0,0) (5.1 , 20.4) (10.2 , 0)
did this thingy
0 = 0^2 a + 0 b + c

20.4 = 5.1^2 a + 5.1 b + c
simplified
20.4 26 a + 5.1 b + c

10.2 = 10.2^2 a + 10.2 b + c
simplified
10.2 = 104a + 10.2 b + c

then did matrix thingy in my calculator
matrix a =
[0 0 1]
[26 5.1 1]
[104 10.2 1]

and set up matrix b after it
[5]
[25.7]
[15.2]

i set up a then b next to each other and asked my calcullator to solve for variables just by pressing enter and got this

[[15.2]
[276]
[797]

so i pluged those numbers into the equation for a quadratic and got
y = 15.2X^2 + 276X + 797

clrealy missed up

i read above posts and am lost

looks like i missed up put still wrong answer

so i corrected matrix b

[0]
[25.7]
[15.2]
yada yada

and got this
[[15.2]
[146]
[227]

still increadibly wrong
y = 15.2 X^2 + 146X = 227

The Lagrange method that the Count gave you is super for this problem because of the zeros in your points:

y = Y1 [(x-x2)(x-x3)]/[(x1-x2)(x1-x3)]
+Y2[(x-x1)(x-x3)]/[(x2-x1)(x2-x3)]
+Y3[(x-x1)(x-x2)]/[(x3-x1)(x3-x2)]
so
y = 0
+20.4[(x)(x-10.2)]/[(5.1)(5.1-10.2)]
+0
so
y = 20.4[x^2 - 10.2 x] /-26.01
y = -.770 x^2 + 8 x
Should carry more places than simply -.770 but that checks

look at your 3 points

x y
0 0
5.1 20.4
10.2 0
the axis of symmetry is at x = 5.1 and the vertex is at (5.1,20.4) because 5.1 is half way between the zeros
4 a (y-k) = (x-h)^2
is the general form for a parabola with vertex at (h,k). In this case a will be negative because this opens down
so
4 a (y -20.4) = (x-5.1)^2
when x = 0, y = 0 so
4 a (-20.4) = 26.01
a = - .31875
so
-1.275 (y-20.4) = x^2 -10.2 x + 26.01
-1.275 y + 26.01 = x^2 - 10.2 x + 26.01
y = -.78431 x^2 + 8 x

The way Count Iblis showed you is the cool way to do it.

It seems like you're trying to find the equation of the parabolic path of a projectile that is launched with an initial velocity of 10.0 m/s at an angle of 45.0 degrees. You've gathered three points from this projectile's trajectory: (0,0), (5.1, 20.4), and (10.2, 0).

To find the equation of a parabola, you need three points or the vertex point and the equation of the axis of symmetry. In this case, you have three points, so you can proceed with finding the equation of the parabola.

One way to find the equation of a parabola is to use the general formula y = ax^2 + bx + c, where a, b, and c are the constants that need to be determined.

To find these constants, you can substitute the coordinates of each point into the equation and form a system of equations. For example, for the point (0,0), you would have the equation 0 = a(0)^2 + b(0) + c, which simplifies to c = 0. Similarly, for the point (5.1, 20.4), you would have the equation 20.4 = a(5.1)^2 + b(5.1) + c, and for the point (10.2, 0), you would have the equation 0 = a(10.2)^2 + b(10.2) + c.

To solve this system of equations, you can use matrix methods or techniques such as substitution or elimination. It appears that you used matrix methods and found the values for a, b, and c, which are 15.2, 276, and 797 respectively.

Therefore, the equation of the parabola that represents the projectile's trajectory is:

y = 15.2X^2 + 276X + 797.

So, that's what gives!