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Mathematics 17 Online
OpenStudy (anonymous):

PLEASE PLEASE HELP ME !!! if the center of the target is 4 feet above the ground and 120 feet from the archer and if the bull's eye is 9.6 inches in diameter. Why does it equal .35 ?

OpenStudy (tkhunny):

Why does what equal 0.35? Is there a question?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

what is the mathematical explanation

OpenStudy (anonymous):

If the center of the target is 4 feet above the ground and 120 feet from the archer, and if the bull’s eye is 9.6 inches in diameter, does the archer score a bull’s eye? Does the arrow land above or below the center of the target; above, below, or within the bull’s eye area? Give mathematical evidence for your answer and assume that the arrow does not veer left or right.

OpenStudy (tkhunny):

There isn't an explanation. So far, it makes no sense. We need more information about the arrow. Is there an initial velocity of the arrow? Is there an initial height of the arrow? Is there a launch angle for the arrow?

OpenStudy (anonymous):

Incomplete question.

OpenStudy (anonymous):

h(x)=-0.00033x^2 +0.036x+4.75

OpenStudy (anonymous):

That's the equation in my book

OpenStudy (anonymous):

No velocity is given

OpenStudy (nottim):

Is there a diagram? Or is the constant for the bow's tension given in a constants page?

OpenStudy (nottim):

Does it ever say to refer to a certain diagram, section or question?

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

it just gives that equation and says to figure out where it hits the target

OpenStudy (nottim):

you're going to have to skip this question; there's no other way unless you can refer to something else for info. You're going to have to ask a teacher.

OpenStudy (anonymous):

ok thank you !

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