An archer releases an arrow from a shoulder height of 1.39 m. When

Question

the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the
arrow along its parabolic path. please help!!I know my points are (0,1.39),(18, 1.684),(45,0) but I don't know how to put them in a quadratic equation.

Answer 1

let the equation be
y = ax^2 + bx + c
you have 3 points
(0,1.39) ---> 1.39 = 0 + 0 + c , ahh c = 1.39
(18,1.684) --> 1.684 = 324a + 18b + 1.39
324a + 18b = .258 --> 162a + 9b = .129 , #1
(45,0) ----->2025a + 45b + 1.39 = 0
2025a + 45b = -1.39
405a + 9b = -.278 , #2

#2 - #1 ---> 243a = -.407
a = -.001675
b = .0448
c = 1.39

y = -.001675x^2 + .0448x + 1.39 (rounded off a bit)

btw, Wolfram agrees with my answer
http://www.wolframalpha.com/input/?i=quadratic+through+%280%2C1.39%29%2C%2818%2C1.684%29%2C+%2845%2C0%29

comment: I don't know where you got those points from, but all 3 satisfy the equation I found
I was expecting "a" to be -4.9 according to gravity

Answer 2

We know the trajectory is along the parabola

y = 1.39 + ax + bx^2

and that

y(18) = 1.684
y(45) = 0

So,

1.39 + 18a + 324b = 1.684
1.39 + 45a + 2025b = 0

a = 0.0478
b = -.00175

y = -.00175x^2 + .0478x + 1.39
max altitude is reached at (13.657,1.716)

Answer 3

Thank u...But how did u get the 162a+9b=.129??

Answer 4

divided through by 2, just to get 9b for eliminating with #2.

algebra I, babe.