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heathernelly Group Title

can someone please

  • 2 years ago
  • 2 years ago

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  1. him1618 Group Title
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    have u done parametric equations?

    • 2 years ago
  2. him1618 Group Title
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    and do you know cartesian coordinates?

    • 2 years ago
  3. heathernelly Group Title
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    no:/

    • 2 years ago
  4. him1618 Group Title
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    lets take the feet of the archer to be (0,0)

    • 2 years ago
  5. heathernelly Group Title
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    these are the points i got (0,1.39) (18,8) (45,0) :/

    • 2 years ago
  6. him1618 Group Title
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    very good..now we need an equation which relates x and y dont we?

    • 2 years ago
  7. heathernelly Group Title
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    yeaah

    • 2 years ago
  8. him1618 Group Title
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    lets say he launches the arrow with a horizontal velocity of v..the vertical velocity initaially is zero

    • 2 years ago
  9. him1618 Group Title
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    can u tell me the x-coordinate at any point in time?

    • 2 years ago
  10. heathernelly Group Title
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    hmm, i don't know :/ ? im not very good at math.. ha

    • 2 years ago
  11. him1618 Group Title
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    distance is speed x time..since there is no acceleration in the x-direction, the speed always remains v..so x = vt

    • 2 years ago
  12. him1618 Group Title
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    understand?

    • 2 years ago
  13. heathernelly Group Title
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    yeah,, kindaa

    • 2 years ago
  14. him1618 Group Title
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    similarly, there is a downward acceleration of -10 m/s^2 in the y-direction.. in the presence of accn, we have the distance covered as -0.5at^2, where a=accn since we are starting at y=1.39 m, we remodel y as 1.39 - 0.5at^2, a= 10 therefore y= 1.39 - 5t^2

    • 2 years ago
  15. heathernelly Group Title
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    my teacher said that i would have to use quadratic equation to do this problem thou ..does this lead to that orr? :/

    • 2 years ago
  16. phi Group Title
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    I posted an answer to this same question (but a different user posted it) But the problem is I don't believe the height of A. You have it as 8 cm. In the real world, this problem does not make sense. Are you sure about the height?

    • 2 years ago
  17. heathernelly Group Title
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    yeah, there's a diagram , i forgot what their called .. but next to the bulls eye it says A (8cm)

    • 2 years ago
  18. heathernelly Group Title
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    my teacher also said The distance from the top edge of the archery target to the ground is 1.5 meters

    • 2 years ago
  19. phi Group Title
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    Here is the other post http://openstudy.com/users/phi#/updates/507e201be4b0919a3cf31396

    • 2 years ago
  20. phi Group Title
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    The distance from the top edge of the archery target to the ground is 1.5 meters That is very important. Is the picture the same as in the post I just listed?

    • 2 years ago
  21. heathernelly Group Title
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    yess, it is

    • 2 years ago
  22. phi Group Title
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    so you use the picture to find the height of A above the ground |dw:1350594969329:dw|

    • 2 years ago
  23. phi Group Title
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    In my post I just assumed A was at (18,1.35) however, based on your new info, A is at (18,1.26) so the numbers have to be re-done.

    • 2 years ago
  24. phi Group Title
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    I did not explain where the equations come from (it's physics), so it is just showing the math

    • 2 years ago
  25. heathernelly Group Title
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    so all i need to do now would be to use the equation ax^2+bx+c ? & then put the #'s in ?

    • 2 years ago
  26. phi Group Title
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    did you find my write-up (it's an attachment in the 2nd to last post)

    • 2 years ago
  27. heathernelly Group Title
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    yes got it, you are awesomee! <33 i don't know what i would of donee. haha

    • 2 years ago
  28. heathernelly Group Title
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    where did you get the -4.9 from?

    • 2 years ago
  29. phi Group Title
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    I used g= 9.8 m/s^2 in 1/2 g t^2

    • 2 years ago
  30. phi Group Title
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    you have to re-do the numbers. And I notice a typo... but it all changes anyway...

    • 2 years ago
  31. heathernelly Group Title
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    is the 1.035 the mistake?

    • 2 years ago
  32. phi Group Title
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    yes, that should have been 1.35 but now it should be 1.26

    • 2 years ago
  33. heathernelly Group Title
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    yeahh :) 18tanθ + 0.04 where did the 0.04 come from?

    • 2 years ago
  34. heathernelly Group Title
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    nevermind, i know now (:

    • 2 years ago
  35. phi Group Title
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    let me know what you get, and I'll double check it.

    • 2 years ago
  36. heathernelly Group Title
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    what does that zero thing mean after tan ? −18tanθ

    • 2 years ago
  37. phi Group Title
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    which equation number are you looking at?

    • 2 years ago
  38. heathernelly Group Title
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    0 =−18tanθ − 0.04 /18^2 45^2 + 45tanθ +1.39

    • 2 years ago
  39. phi Group Title
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    I used the point (45,0) in the equation (the arrow hits the ground y=0 45 meters away from the archer x=45 so that 0 is y

    • 2 years ago
  40. phi Group Title
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    I labeled this equation (6) so you can refer to it.

    • 2 years ago
  41. heathernelly Group Title
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    i mean like how did you get this - tanθ =19/1125 = 0.0168

    • 2 years ago
  42. phi Group Title
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    solve for tanθ it is a bit messy, and I did not bother to write it all down. It is algebra. You do know algebra :)

    • 2 years ago
  43. heathernelly Group Title
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    haha yeah, wasnt the best at it but i know what tan is ..so i would type in my calculator -18tan ___? what goes in the blank thou :/

    • 2 years ago
  44. phi Group Title
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    are you working with the new numbers, or just working through the numbers in this old post?

    • 2 years ago
  45. heathernelly Group Title
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    new numbers, haha

    • 2 years ago
  46. phi Group Title
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    what equation do you have now?

    • 2 years ago
  47. heathernelly Group Title
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    i have 0 = 18tan - 0.13 / 18^2? :/

    • 2 years ago
  48. phi Group Title
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    what about the rest? I would expect to see \[ y= \frac{18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39\]

    • 2 years ago
  49. heathernelly Group Title
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    yeaah i have it, i forgot to say that

    • 2 years ago
  50. heathernelly Group Title
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    i actually have for the rest 45^2 + 45 tan + 1.39

    • 2 years ago
  51. phi Group Title
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    yes, after you sub in (45,0) you get \[ 0= \frac{18\tan(\theta)-0.13}{18^2}45^2+45 \tan(\theta)+1.39\]

    • 2 years ago
  52. heathernelly Group Title
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    yeah :) but then i get confused after that bit because i dont know how you got 19/1125

    • 2 years ago
  53. phi Group Title
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    get out a piece of paper and use the distributive property and re-write the equation \[ 0= \frac{-45^2\tan(\theta)}{18} -\frac{0.13}{18^2}45^2+45 \tan(\theta)+1.39\] I divided everything by 45: \[ 0= -\frac{45}{18}\tan(\theta) -\frac{0.13}{18^2}45+ \tan(\theta)+\frac{1.39}{45}\] the 45/18 simplifies to 5/2. I moved the tan terms together \[ 0= (1-\frac{5}{2})\tan(\theta) -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\] you can finish I hope

    • 2 years ago
  54. heathernelly Group Title
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    yes, thank you soo soo much for the help :)

    • 2 years ago
  55. phi Group Title
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    post your results. I'll check back laster.

    • 2 years ago
  56. heathernelly Group Title
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    will do :)

    • 2 years ago
  57. TiffanyLee3 Group Title
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    Has anyone figured this out?

    • 2 years ago
  58. heathernelly Group Title
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    not yet, still trying to figure it out :)

    • 2 years ago
  59. heathernelly Group Title
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    i figure out that it would be 0.0128333 ..so y = ____x^2 + 0.0128333+1.39 but i can't figure out what the other # is :/

    • 2 years ago
  60. phi Group Title
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    how did you get 0.01283... ?

    • 2 years ago
  61. heathernelly Group Title
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    from 0=(1-5/2)tan - 0.13/18 x 5/2 + 1.39/45 :/

    • 2 years ago
  62. phi Group Title
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    you start with \[ 0= (1-\frac{5}{2})\tan(\theta) -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\] 1-5/2 is 2/2 -5/2= -3/2 \[0=-\frac{3}{2}\tan(\theta) -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\] or \[\frac{3}{2}\tan(\theta) = -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\] the stuff on the right simplifies to 77/6000 (which is 0.01283...) \[\frac{3}{2}\tan(\theta) = \frac{77}{6000} \] and \[\tan(\theta) = \frac{77}{6000}\cdot \frac{2}{3}\]

    • 2 years ago
  63. heathernelly Group Title
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    so 0.085

    • 2 years ago
  64. phi Group Title
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    so tanθ = 77/9000 = 0.008555... that is your coefficient on x you can find the coefficient of the x^2 term by replacing tanθ

    • 2 years ago
  65. phi Group Title
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    I think you missed a decimal place

    • 2 years ago
  66. heathernelly Group Title
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    yeah, haha ..sorry 0.0085 ..so i would use that & the number i said up there^^?

    • 2 years ago
  67. phi Group Title
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    \[ y= \frac{-18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39 \] you need to find the coeff for the x^2 term. but you know tan so you can simplify

    • 2 years ago
  68. heathernelly Group Title
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    i just got kinda confused :(

    • 2 years ago
  69. phi Group Title
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    Yes, this is a complicated problem, and you have to be methodical. If you look at my pdf, we got to equation (5). Then we sub in (45,0) and solved for tan= 77/9000 now you can replace tanθ in equation (5): \[ y= \frac{-18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39 \] you will get a parabola with numbers for coefficients

    • 2 years ago
  70. phi Group Title
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    just get out your calculator or type it in google: (-18*77/9000 -0.13)/(18^2)=

    • 2 years ago
  71. heathernelly Group Title
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    -0009765

    • 2 years ago
  72. phi Group Title
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    oooh, Detroit swept the Yankees 4 games to 0. Detroit is off to the World Series

    • 2 years ago
  73. heathernelly Group Title
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    haha, i dont watch that ..but woo for them! :D

    • 2 years ago
  74. phi Group Title
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    you have to be more careful. put in a decimal point, and it is not 9 for the first digit

    • 2 years ago
  75. heathernelly Group Title
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    8 i meaan (:

    • 2 years ago
  76. heathernelly Group Title
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    i mean -0.0008765

    • 2 years ago
  77. phi Group Title
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    so now you have the full equation.

    • 2 years ago
  78. heathernelly Group Title
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    yay ..so do i use the 0.0085 & that?:)

    • 2 years ago
  79. phi Group Title
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    yes

    • 2 years ago
  80. heathernelly Group Title
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    omg, i would of never done this without you

    • 2 years ago
  81. phi Group Title
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    you still have to find the max height of y

    • 2 years ago
  82. heathernelly Group Title
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    yeahh , so i did that next step ..which i got -3.725

    • 2 years ago
  83. phi Group Title
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    The next step is to find x where the parabola peaks. The way the problem is set up, the archer is at x=0 and the target at x=18 x= -3.7.. is behind the archer. So maybe you got the wrong sign?

    • 2 years ago
  84. phi Group Title
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    How did you find x?

    • 2 years ago
  85. heathernelly Group Title
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    how did i get -3.725? :/

    • 2 years ago
  86. phi Group Title
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    if you have y = ax^2 +bx+c you should use x_max= -b/(2a)

    • 2 years ago
  87. heathernelly Group Title
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    yes i used -b/2a to get the answer -3.7

    • 2 years ago
  88. heathernelly Group Title
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    25

    • 2 years ago
  89. phi Group Title
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    maybe you should type out the equation y = -0.0008765 x^2 + 0.0085 x + 1.39

    • 2 years ago
  90. heathernelly Group Title
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    yess, i wrote that down on my paper :) haha

    • 2 years ago
  91. heathernelly Group Title
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    but i had it the wrong way around :/ oh

    • 2 years ago
  92. phi Group Title
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    yes, I notice you are not careful enough.

    • 2 years ago
  93. heathernelly Group Title
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    im confused on what to do next

    • 2 years ago
  94. phi Group Title
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    y = -0.0008765 x^2 + 0.0085 x + 1.39 type in google - 0.0085/(2*-0.0008765)=

    • 2 years ago
  95. heathernelly Group Title
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    4.8488

    • 2 years ago
  96. phi Group Title
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    much better. that is the x value where the parabola peaks. what is the y value at that x value?

    • 2 years ago
  97. phi Group Title
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    type this in google -0.0008765*4.8488^2 + 0.0085*4.8488 + 1.39=

    • 2 years ago
  98. heathernelly Group Title
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    1.41

    • 2 years ago
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