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.
My points are
(0,1.39)
(18,1.5)
(45,0)
I'm supposed to use this formula: y = ax^2 + bx + c
I get C = 1.39 after that I get lost

Answer 1

ax^2+bx+c=y

0a+0b+c=1.39
324a+18b+c=1.5
2025a+45b+c=0

a=-.00137
b=0.0308
c=1.39

y = -.00137x^2 + 0.031x + 1.39

max height on the parabola is at x = -b/2a, so at
x = .031/.00274 = 11.31
y = 1.565