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BreezeFlow

  • 2 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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  1. theEric
    • 2 years ago
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    Hi! 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!

  2. BreezeFlow
    • 2 years ago
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    Thank 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.

  3. Algebraic!
    • 2 years ago
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    this question has been posted before. there's not enough info here to reach a conclusion without making an assumption...

  4. BreezeFlow
    • 2 years ago
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    Can you see the picture?

  5. Algebraic!
    • 2 years ago
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    "From top to bottom the whole target stand is supposed to be 1.5 meters." where does it say that??

  6. BreezeFlow
    • 2 years ago
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    On 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

  7. Algebraic!
    • 2 years ago
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    then it's easy to solve.

  8. theEric
    • 2 years ago
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    Haha, that might be a necessary piece of information! It makes point A known! :)

  9. BreezeFlow
    • 2 years ago
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    Oh good it can be solved :D

  10. theEric
    • 2 years ago
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    And I was silly to think the shot was horizontal - the answer would be shoulder height :P Do you want to take it away, @Algebraic! ?

  11. Algebraic!
    • 2 years ago
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    not 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...

  12. BreezeFlow
    • 2 years ago
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    IS 1.1 the height of A?

  13. Algebraic!
    • 2 years ago
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    OS crashed the pdf page... what were the measurements...

  14. Algebraic!
    • 2 years ago
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    1.5 - the distance from the top... whatever it was something like 4cm * 4 + 8cm?

  15. theEric
    • 2 years ago
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    The height was 1.5m, and the point was 24 cm down.

  16. theEric
    • 2 years ago
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    .24m down.

  17. BreezeFlow
    • 2 years ago
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    Here'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

  18. Algebraic!
    • 2 years ago
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    ah ok so (18, 1.26) is the second point...

  19. BreezeFlow
    • 2 years ago
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    Do I put the points into the quadratic equation form?

  20. BreezeFlow
    • 2 years ago
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    When I was working on this earlier I put c as being 1.39 for one of the equations.

  21. Algebraic!
    • 2 years ago
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    I'd use... y= a(x-h)^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...

  22. Algebraic!
    • 2 years ago
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    1.26 = a(18)^2 +b18 +1.39 0 = a(45)^2 +b(45) +1.39

  23. Algebraic!
    • 2 years ago
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    a is obviously going to be negative..

  24. BreezeFlow
    • 2 years ago
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    Do i solve it like linear systems?

  25. Algebraic!
    • 2 years ago
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    yep

  26. BreezeFlow
    • 2 years ago
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    Can I work this out and you guys tell me if i did it right?

  27. Algebraic!
    • 2 years ago
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    sure

  28. BreezeFlow
    • 2 years ago
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    Ok, thank you I'll get working on it.

  29. BreezeFlow
    • 2 years ago
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    Ok 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?

  30. BreezeFlow
    • 2 years ago
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    From 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?

  31. Algebraic!
    • 2 years ago
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    no.

  32. BreezeFlow
    • 2 years ago
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    What should I do instead?

  33. Algebraic!
    • 2 years ago
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    I solved -.13 = 18^2a +18b for b and subsed in to the other equation a~ -8.765E-4 b~ .008555

  34. BreezeFlow
    • 2 years ago
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    To find the maximum height do I use the formula -b/2a?

  35. Algebraic!
    • 2 years ago
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    that's the x location of the max height... you plug that into the equation to find the actual height...

  36. BreezeFlow
    • 2 years ago
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    What's the equation for height. Oh and with the number for a should i type e-4 in my calculator?

  37. Algebraic!
    • 2 years ago
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    10^-4

  38. Algebraic!
    • 2 years ago
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    we just found the equation for height...

  39. BreezeFlow
    • 2 years ago
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    We use the second equation?

  40. Algebraic!
    • 2 years ago
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    the equation of the parabola... we were using the points... to find the equation of the parabola to find its vertex....

  41. BreezeFlow
    • 2 years ago
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    I 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

  42. Algebraic!
    • 2 years ago
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    the vertex is the location of the max height

  43. Algebraic!
    • 2 years ago
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    |dw:1350531626084:dw|

  44. BreezeFlow
    • 2 years ago
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    Does that mean the y-coordinate of the vertex is the max height?

  45. Algebraic!
    • 2 years ago
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    (x,y) = (x location of max height, max height)

  46. Algebraic!
    • 2 years ago
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    yes

  47. BreezeFlow
    • 2 years ago
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    Do i already know the vertex? Was it one of the points from earlier?

  48. Algebraic!
    • 2 years ago
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    yeah it was (45,0)

  49. Algebraic!
    • 2 years ago
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    you already knew the answer before you did the problem, so this was all a huge waste of time.

  50. BreezeFlow
    • 2 years ago
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    I still don't get how to arrive to the answer and I need to show the work

  51. Algebraic!
    • 2 years ago
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    find a and b y= ax^2 +bx +c find the vertex

  52. BreezeFlow
    • 2 years ago
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    If the vertex is supposed to be (45,0) wouldn't that make the maximum height 0? o_o

  53. Algebraic!
    • 2 years ago
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    (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

  54. BreezeFlow
    • 2 years ago
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    When I try to find the height I get -4.88025002715, but I know that can't be write.

  55. Algebraic!
    • 2 years ago
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    it's not.

  56. BreezeFlow
    • 2 years ago
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    I 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 y-coordinate.

  57. Algebraic!
    • 2 years ago
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    plug x =-b/2a into the equation of the parabola

  58. BreezeFlow
    • 2 years ago
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    The ax^2+bx+c type equation?

  59. Algebraic!
    • 2 years ago
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    alternately, calculate (4ac -b^2) / 4a your choice.

  60. BreezeFlow
    • 2 years ago
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    Ok I'll try that

  61. BreezeFlow
    • 2 years ago
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    I got 1.3691, but I probably miscalculated somewhere unless I'm supposed to round it to 1.4

  62. Algebraic!
    • 2 years ago
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    it's 1.41

  63. Algebraic!
    • 2 years ago
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    1.410876 using the rough values for a and b that I gave

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