If I know two points, how can I calculate a parabola?

Question

I'm trying to calculate the path of a snowball that is starting at (1,1) and arcing over to (5,3) Any thoughts on how to write out the equation?

I'm trying to calculate the path of a snowball that is starting at (1,1) and arcing over to (5,3) Any thoughts on how to write out the equation?
I want to be able to generate random parabolas based on teh x of the vertex being somewhere between the to points, i.e. the vertex would be (3,Yv) How do I solve for Yv?

Answer 1

There are many parabolas that go through those points. For example if you want one of the form y=ax^2+bx,you can substitute in (1,1) and get a+b=1 and then substitute in (5,3) and get 25a+5b=3. Solving a=-1/10 and b=11/10. This gives a parabola y=-1/10x^2+11/10x which goes through the given points, but it is one of many. Any more constraints?

Answer 2

Parabola is y = ax² + bx + c. 2 points don't determine a parabola uniquely unless you know one of the coefficients, or unless one point is the vertex. In this case, the a coefficient is -16 ft/s² if the snowball is on this planet. So plugging in the points gives us

1 = -16(1)² + b(1) + c
3 = -16(5)² + b(5) + c

subtracting gives

2 = -16(24) + 4b
4b = 386
b = 96.5

then get c from
3 = -16(5)² + 96.5(5) + c
3 = -400 + 482.5 + c
c = -79.5

which isn't realistic, since c is the starting height. Without a 3rd point or other special information, this is the best you can do.

Answer 3

the equation of parabola is A x^2 + B x + C = 0
there are 3 unknown A ,B & C
You need 3 points to substitute to solve this problem
lacking one point

Answer 4

Actually, there are an infinite number of parabolas that can pass through those two points. You need either a third point or one of the coefficients in the polynomial in order to have a unique solution.

Answer 5

As stated above you would either need a 3rd point, or the acceleration of the snowball. Are you assuming that it is carried downward by gravity at 9.8 m/s²?