Question

Please check my work and let me know what I did wrong.
An archer releases an arrow from a shoulder height of 1.39m. When the arrow hits the target 18m away, it hits point A. When the target is removed, the arrow lands 45m away. Find the maximum height of the arrow along its parabolic path.

Answer 1

This is what I have and was toActivity 2; Analyzing
Points are; f(0)=1.39
f(18)=h=1.26
f(45)=0
(0,139),(18,1.26),(45,0)
Using the equation y=f(x)=ax^2+bc+c, where c is 1.39
a(0)^2+b(0)+c=1.39, c=1.39
a(18)^2+b(18)+c=h
a(45)^2+b(45)+c=o
324a+18b=h-1.39
2025a+450b=-1.39
Vertex of the graph is the maximum height when it is (45,0)
To find the height plug x=-b/2a in the equation of the parabola, (4ac-b^2)/4a, estimating the solution 1.410876, making the answer 1.41

Activity 3; Modeling
The points are (140,1.4)(150,1.25)(170,0.93)(175,0.78)(205,0.43), by using the online regression calculator, I’m able to find which models are better suited for the points of the graph. The linear regression line equation is estimated to 3.513-0.015x. The quadratic regression line equation is y=0.0001x^2-0.0348+5.1717. I wasn’t able to use a graphing calculator so I used the substituting points into each equation. By this I found the linear equation to be the one that would be better suited.
3.513-0.015x=(140,1.41)(150,1.26)(170,0.96)(175,0.89)(205,0.44)
Y=0.0001x^2-0.0348+5.1717=(140,2.26)(150,2.2)(170,2.15)(175,2.14)(205,2.24)
The quadratic line appears a little higher than that of the linear, and the points are closer to the linear line. So I believe that the linear equation is better suited.
ld it's wrong.

Answer 2

Please someone help explain. I know the question has been asked a hundred times, but I just don't get it