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anonymous
 3 years ago
I would really appreciate if someone could help me with this problem. It relates parabolas and archery. I've been trying to figure it out for about two hours now and I can't seem to get to the right answer, which I already know. Here's a link so you can see the picture with the problem:
http://mrswhaples.cmswiki.wikispaces.net/file/view/Algebra+II+chapter+4+solutions.pdf
The problem I need help with is on the second last page under Activity 2. Every time I attempt to get to the right answer, I am way off. D:
anonymous
 3 years ago
I would really appreciate if someone could help me with this problem. It relates parabolas and archery. I've been trying to figure it out for about two hours now and I can't seem to get to the right answer, which I already know. Here's a link so you can see the picture with the problem: http://mrswhaples.cmswiki.wikispaces.net/file/view/Algebra+II+chapter+4+solutions.pdf The problem I need help with is on the second last page under Activity 2. Every time I attempt to get to the right answer, I am way off. D:

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theEric
 3 years ago
Best ResponseYou've already chosen the best response.0Hi! I want to help, but I'm not quite sure how this is working. So maybe we can work together to get to the right answer. You can tell me what you did and I'll try to understand it, or we can start from the beginning. Is it assumed that the archer shoots horizontally? And how high is the target? Are we able to know that? Sometimes I think we'd need to know things that we really don't. Variables cancel out, sometimes.. Also, I'm sorry you've spent so much time on this so far!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for being willing to help. So far I've been using three points to try to find an answer: (0, 1.39) (18,40) and (45,0). From top to bottom the whole target stand is supposed to be 1.5 meters. And the arrow is being shot at 1.39 meters.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this question has been posted before. there's not enough info here to reach a conclusion without making an assumption...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you see the picture?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"From top to bottom the whole target stand is supposed to be 1.5 meters." where does it say that??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0On my homework sheet it says that. It's supposed to be a tip. Use the following piece of information to help you solve Activity 2: The distance from the top edge of the archery target to the ground is 1.5 meters

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then it's easy to solve.

theEric
 3 years ago
Best ResponseYou've already chosen the best response.0Haha, that might be a necessary piece of information! It makes point A known! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh good it can be solved :D

theEric
 3 years ago
Best ResponseYou've already chosen the best response.0And I was silly to think the shot was horizontal  the answer would be shoulder height :P Do you want to take it away, @Algebraic! ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not really... you have three points now. (0, 1.39) (18, 1.1) and (45,0) should not be a problem to find the vertex of that parabola...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0IS 1.1 the height of A?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OS crashed the pdf page... what were the measurements...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01.5  the distance from the top... whatever it was something like 4cm * 4 + 8cm?

theEric
 3 years ago
Best ResponseYou've already chosen the best response.0The height was 1.5m, and the point was 24 cm down.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here's the typed out problem 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. 4 cm 4 cm 4 cm 4 cm 8 cm

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah ok so (18, 1.26) is the second point...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do I put the points into the quadratic equation form?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When I was working on this earlier I put c as being 1.39 for one of the equations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'd use... y= a(xh)^2 +k for ease... but yeah you can use y = ax^2 +bx +c with c=1.39m if you like... doesn't really matter...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01.26 = a(18)^2 +b18 +1.39 0 = a(45)^2 +b(45) +1.39

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a is obviously going to be negative..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do i solve it like linear systems?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can I work this out and you guys tell me if i did it right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, thank you I'll get working on it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok so far I simplified the expressions to 324a + 18b +1.39= 1.26 and 2025a+45b+1.39=0 Do I then subtract 1.39 from both sides and use the method of multiplying on equation by a number so that the coefficients of one variable will be opposites?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From subtracting 1.39 from both sides I got 324a + 18b = 0.13 and 2025a+45b=1.39 Now, should I multiply one equation like 324a + 18b = 0.13 by 2.5 so the coefficient of b will be 45?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What should I do instead?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I solved .13 = 18^2a +18b for b and subsed in to the other equation a~ 8.765E4 b~ .008555

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To find the maximum height do I use the formula b/2a?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the x location of the max height... you plug that into the equation to find the actual height...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What's the equation for height. Oh and with the number for a should i type e4 in my calculator?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we just found the equation for height...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We use the second equation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the equation of the parabola... we were using the points... to find the equation of the parabola to find its vertex....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have to find the vertex before I find the height? Sorry, if I sound stupid, I've been really confused with this problem for awhile. XP

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the vertex is the location of the max height

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350531626084:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does that mean the ycoordinate of the vertex is the max height?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(x,y) = (x location of max height, max height)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do i already know the vertex? Was it one of the points from earlier?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you already knew the answer before you did the problem, so this was all a huge waste of time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I still don't get how to arrive to the answer and I need to show the work

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0find a and b y= ax^2 +bx +c find the vertex

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If the vertex is supposed to be (45,0) wouldn't that make the maximum height 0? o_o

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(0, 1.39) was a point so y= a(0)^2 +b(0) +c 1.39 =c (18, 1.26) was a point so 1.26 = a(18^2) + b(18) +1.39 (45,0) was a point so 0 = a(45)^2 +b(45)+ 1.39

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When I try to find the height I get 4.88025002715, but I know that can't be write.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know how the answer is supposed to be 1.41 and I don't know how to find the vertex that has 1.41 as the ycoordinate.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plug x =b/2a into the equation of the parabola

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The ax^2+bx+c type equation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alternately, calculate (4ac b^2) / 4a your choice.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got 1.3691, but I probably miscalculated somewhere unless I'm supposed to round it to 1.4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01.410876 using the rough values for a and b that I gave
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