Question

An archer releases an arrow from a shoulder height of 1.39m away, it hits point A. When the target is removed, the arrow lands 45m away. Find the maximun height of the arrow along its parabolic path. How am I supposed to solve the equation to find the max height.

Answer 1

For the first, the parabola is defined by the equation:

y = f(x) = ax² + bx + c

...where a, b, and c are constant coefficients to be determined. You are given three points:

f(0) = 1.39
f(18) = h ... (height of point A) ... I suspect there's a figure you're not showing us
f(45) = 0

You need that height h to solve the problem. That will give you three equations in 3 unknowns:

a(0²) + b(0) + c = 1.39 ... this makes c = 1.39 no matter what h is
a(18²) + b(18) + c = h
a(45²) + b(45) + c = 0

You can simplify that to two equations, since c=1.39 is known:
324a + 18b = h - 1.39
2025a + 45b = -1.39

When you get a value for h, you can solve for a and b. If a is not negative, you have a problem since a trajectory should always be a parabola that opens downward. The maximum height will be where x = -b/(2a), and the height will be
f(-b/(2a)) = a(-b/(2a))² + b(-b/(2a)) + c
= b²/(4a) - b²/(2a) + c = c - b²/(4a)

Since b² is positive and a is negative, that will be greater than c, as expected.

2. Your calculator (if it has linear regression) will give you a model that look like;

y = am + b ... m = mass added, y = amount of bend

...for the linear model, with a and b as the coefficients of the linear model. b should be nearly 0 in this case if it's a good fit. Just plug in the (mass, bend distance) values for the (independent, dependent) variables and read out the coefficients a and b at the end.

Typically, you only get one-variable linear regression, so the best quadratic model you can solve with that tool looks like:

y = am² + c ... same y and m, but a and c are the coefficients this time

For that, plug the squares of the masses, instead, as the independent variables. Then you will get the least-squares solution for the a and c coefficients. If you have a two-variable regression option that solves for a model a(x1) + b(x2) + c = y, then you can use m² for x1, m for x2 and get the a, b and c value for a model of y = am² + bm + c

Answer 2

Medal?

Answer 3

Skip the #2 that is extra info.

Answer 4

Okay so I'm stuck at the 324a+18b=-1.39 and the other as well what am I supposed to do after that I was looking at the same thing on wiki answers that's where I am now.