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Mathematics 14 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

@satellite73 Can you help me with this question?

OpenStudy (hba):

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

OpenStudy (hba):

You need that height h to solve the problem.

OpenStudy (hba):

That will give you three equations in 3 unknowns

OpenStudy (anonymous):

Oh, the height is suppose to be 5 feet.

OpenStudy (hba):

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)

OpenStudy (anonymous):

okay, i am confused.

OpenStudy (hba):

Till Where Did You Really Get It ?

OpenStudy (anonymous):

Well, I understand the equation, not really when you labeled the 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"

OpenStudy (anonymous):

@hba

OpenStudy (hba):

Okay So Let Me Continue From Here :)

OpenStudy (hba):

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 . 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

OpenStudy (hba):

ax² + bx + c ^^

OpenStudy (hba):

You Get It ? @TiffanyLee3

OpenStudy (anonymous):

yea, a little, I was just reading over it

OpenStudy (hba):

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)

OpenStudy (anonymous):

I have no idea where to even start with that..

OpenStudy (hba):

Its f(x)=ax^2+bx+c=0

OpenStudy (anonymous):

Then what do i do with all of this f(-b/(2a)) = a(-b/(2a))² + b(-b/(2a)) + c = b²/(4a) - b²/(2a) + c = c - b²/(4a)

OpenStudy (hba):

Jus Plug in x= -b/2a

OpenStudy (hba):

I Am Really Sorry But I Have To Leave

OpenStudy (hba):

I hope You Understood ?

OpenStudy (anonymous):

Nope, not really, Ill just leave this one blank

OpenStudy (hba):

No Answer It

OpenStudy (anonymous):

I do not know how.

OpenStudy (anonymous):

Someone else answered the same question earlier. The solution is attached below. All credits to Phi, I take no credit whatsoever. Just restating the answer where it's needed.

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