Project on target. 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 the path.

Question

anonymous · Answer

Note: the distance from the top edge of the target to the ground is 1.5m

anonymous · Answer

@Hero  please hepl

anonymous · Answer

@Hero  I really need the help I dont want the answer I just need guidance in understanding the problem. Thanks!

anonymous · Answer

@jdoe0001 could you give me a hand?

jdoe0001 · Answer

hmmm not sure I know..... since   I"d use the initial velocity formula.... but that'd require the initiali velocity of the arrow.. wince we don't have

jdoe0001 · Answer

which we don't have rather

anonymous · Answer

it says that it wants only the maximum height of the arrows path, do you know any other possible solution that could help me

anonymous · Answer

@jim_thompson5910  could you give me a hand?

jdoe0001 · Answer

am afraid not......
I can tell is a parabola.... and the maximum point would be at its vertex...... I just don't see enough info to get a quadratic equation from that

jim_thompson5910 · Answer

are you given a drawing of this?

anonymous · Answer

i am @jim_thompson5910

anonymous · Answer

http://mastmedical.org/ourpages/auto/2013/12/12/52035199/Chapter%204%20Project.pdf

anonymous · Answer

its under activity two

jim_thompson5910 · Answer

thanks

anonymous · Answer

you will also see that point a is 8cm under the edge of the target which is located at 1.5m

anonymous · Answer

sorry 24cm^

jim_thompson5910 · Answer

I was hoping this drawing would show how high point A is off the ground, but it doesn't mention that at all.

anonymous · Answer

well if the top edge of the target is 1.5m from the ground and point a is 24cm below it is that enough info?

jim_thompson5910 · Answer

where does it say that?

jim_thompson5910 · Answer

that would be enough info, but I don't see that on the PDF

anonymous · Answer

its on tthe image next to the question

jim_thompson5910 · Answer

oh that's not posted?

jim_thompson5910 · Answer

oh wait, one sec

jim_thompson5910 · Answer

nvm I don't see it. I must be missing something

anonymous · Answer

when you look a activity 2 its the image next to the question showing you the distance from the very top of the target to point a

jim_thompson5910 · Answer

yeah that's 24 cm or 0.24 meters (4+4+4+4+8 = 24 cm). I still don't see 1.5 though.

jim_thompson5910 · Answer

anyways, if the top edge of the circle is 1.5 m high off the ground, and this target board is completely vertical, then point A is 1.5 - 0.24 = 1.26 meters off the ground

anonymous · Answer

because 1.5 is given to us in class its a hint. sorry!!!!! i didnt mention that!!!!!

anonymous · Answer

im just stressed an dforgot about it!

jim_thompson5910 · Answer

that's ok. I just wish it was on the PDF because that's very crucial info. Then again, it's definitely an incentive to show up to class lol

jim_thompson5910 · Answer

that's fine

jim_thompson5910 · Answer

let's set up a blank xy axis

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jim_thompson5910 · Answer

then on the y axis, mark 1.39
on the x axis, mark 18 and 45 (the target is at 18 m, the arrow lands on the ground at 45 m without the target there)

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jim_thompson5910 · Answer

at the target, which is at x = 18 m, the arrow hits point A which is 1.26 m off the ground

so we'll have the point (18,1.26)

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