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Mathematics 15 Online
OpenStudy (anonymous):

Medal please help!! 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):

you need more than help with this one. ive got no idea

OpenStudy (anonymous):

tell them its a stupid question. Will Hunting could not even do it.

OpenStudy (anonymous):

haha okay

OpenStudy (anonymous):

is there any more details provided?

OpenStudy (anonymous):

OpenStudy (anonymous):

im only doing 2 & 3

OpenStudy (anonymous):

i just had a look at what u have been given, is there any data that goes along with it? numbers limits etc?

OpenStudy (anonymous):

no lol

OpenStudy (anonymous):

let me think bout it, hope someone else can help if i cant

OpenStudy (anonymous):

okay thanks:)

OpenStudy (anonymous):

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 (anonymous):

You need that height h to solve the problem.

OpenStudy (anonymous):

That will give you three equations in 3 unknowns

OpenStudy (anonymous):

GirlByte is right, we need the height of the of point 'A' when the arrow hits the target 18m away

OpenStudy (anonymous):

girl byte u r too good

OpenStudy (anonymous):

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):

the attachment i posted is all i have :/

OpenStudy (anonymous):

I'd assume the height of point A is the same height as the archer of 1.39m so the three points are: (0, 1.39) (18, 1.39) (45, 0) make up three equations from the quatratic equation of y = ax^2 + bx + c solve for a, b, and c

OpenStudy (anonymous):

Ok. Here's an attachment. This contains the solution. If you have any questions just ask me http://assets.openstudy.com/updates/attachments/508079bae4b067efc9ff1ee0-fall12-13-1350601069012-507e201be4b0919a3cf31396phi1350496607180archer.pdf

OpenStudy (anonymous):

oh i thought is was 1.41

OpenStudy (anonymous):

GirlByte, how do you know the bulls eye is at shoulder height?

OpenStudy (anonymous):

I do archery :)

OpenStudy (anonymous):

what if u r Lebron James and u r 6-8

OpenStudy (anonymous):

do they alter the target for each individual?

OpenStudy (anonymous):

I guess that info is 'general' knowledge in the archery world! ^_^

OpenStudy (anonymous):

Anyways ,back to the topic. @aubreyk9616 Do u have any questions?

OpenStudy (anonymous):

im just confused, i cant figure it out.

OpenStudy (anonymous):

At which point of the solution you find hard to understand? Just ask me ;P

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