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jarp0120

  • one year ago

calculus:rectilinear motion an object is thrown vertically upward from the ground reaches a certain height after 2 sec and returns to the same height on descent, 8 sec later. find its initial velocity and the height in question.

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  1. Jarp0120
    • one year ago
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    |dw:1443848025630:dw|

  2. Jarp0120
    • one year ago
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    is this right?

  3. Miracrown
    • one year ago
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    The upward line is 2s, and the total curve at the top is 8s - That seems right :)

  4. Miracrown
    • one year ago
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    We want to use the idea that acceleration due to gravity is constant to help us Do you know if your class prefers feet or meters? We can use either units, but I don't want to do it in a way you haven't seen =/

  5. Miracrown
    • one year ago
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    @jarp0120

  6. Jarp0120
    • one year ago
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    feet

  7. Jarp0120
    • one year ago
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    and we need to use derivative to find the answer

  8. Miracrown
    • one year ago
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    So an important general law all airborne objects follow is this: d^2 y/dt^2 = -32.2 ft/s^2

  9. Miracrown
    • one year ago
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    You know the second derivative of height is a constant

  10. Jarp0120
    • one year ago
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    so that's that acceleration?

  11. Miracrown
    • one year ago
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    Lets think mostly about the 8 seconds it was highest up first. What do you think the first derivative of its height was when it was at the very top?

  12. Miracrown
    • one year ago
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    The first derivative is velocity. The second derivative is acceleration We know the second derivative is constant

  13. Jarp0120
    • one year ago
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    oh yes okay

  14. Miracrown
    • one year ago
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    Can you answer my question?

  15. Miracrown
    • one year ago
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    If youre not sure yet, that's ok too, its just kind of a shortcut if you realize it

  16. Jarp0120
    • one year ago
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    i don't know

  17. Miracrown
    • one year ago
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    That's ok, lets move forward with the problem a bit and readdress it later, when it comes up. So we have this constant second derivative...But that's not very descriptive. Can you use it to get an equation for the first derivative?

  18. Jarp0120
    • one year ago
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    is the first derivative v=vo-32t?

  19. Miracrown
    • one year ago
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    Yes, that's very well put. :)

  20. Miracrown
    • one year ago
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    So, do you have any ideas how we might find vo? Its ok if you dont, but if you do they could be useful

  21. Jarp0120
    • one year ago
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    i don't really know

  22. Miracrown
    • one year ago
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    Dont worry, even if you cant see a way yet, we can forge ahead blindly and well find something useful eventually :P

  23. Jarp0120
    • one year ago
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    thank you

  24. Miracrown
    • one year ago
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    So often when unsure the best thing to do in real life is just fight forward. So lets do that. We found an equation for velocity. Can you find one for height? (y, or the 0th derivative)

  25. Jarp0120
    • one year ago
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    s=vo-16t^2?

  26. Jarp0120
    • one year ago
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    s= Vot-16t^2?

  27. Jarp0120
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    @miracrown?

  28. Jarp0120
    • one year ago
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    is the length in 2 sec = to the length in 8 sec?

  29. Miracrown
    • one year ago
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    @jarp0120 you said before: s= Vot-16t^2? That's technically correct...did you have a d0 at first, and replace it with zero? or an s0, I guess

  30. Jarp0120
    • one year ago
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    |dw:1443853241921:dw|

  31. Miracrown
    • one year ago
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    So we have an equation for s. But we still dont know vo.Maybe we can figure it out though What two times is the ball at the same height?

  32. Miracrown
    • one year ago
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    @jarp0120

  33. Jarp0120
    • one year ago
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    what?

  34. Miracrown
    • one year ago
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    Did you follow what I said? If not, its ok.

  35. Jarp0120
    • one year ago
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    no hahaha

  36. Miracrown
    • one year ago
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    Well remember how the problem is framed, It takes 2s to go up to a certain point and then 8 more seconds to go up, and back down to that point. So wouldnt it have the same height at t=2s and t=10s?

  37. Jarp0120
    • one year ago
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    yeah

  38. Miracrown
    • one year ago
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    So how about you plug in 2 and 10 into your position equations and set them equal since we know 2 and 10 should give the same result

  39. Jarp0120
    • one year ago
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    yes sir

  40. Miracrown
    • one year ago
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    It was a little bit of a trick there. So, try out my setting equal idea you should be able to use it to solve vo

  41. Jarp0120
    • one year ago
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    what do again in the first derivative?

  42. Miracrown
    • one year ago
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    You want to do it with the position equation

  43. Miracrown
    • one year ago
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    s = vo*t - 16*t^2

  44. Miracrown
    • one year ago
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    you want to set s(2) = s(10)

  45. Jarp0120
    • one year ago
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    s=Vot-16t^2 d=Vo-32t a=-32 right?

  46. Miracrown
    • one year ago
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    s v and a, yes

  47. Miracrown
    • one year ago
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    v = vo - 32t, I think you meant? @jarp0120

  48. Jarp0120
    • one year ago
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    substitute the t?

  49. Jarp0120
    • one year ago
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    yes

  50. Miracrown
    • one year ago
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    yes, you want to sub in 2 and 10 respectively for it

  51. Jarp0120
    • one year ago
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    s(2)=2Vo-64 s(10)=10Vo-1600

  52. Miracrown
    • one year ago
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    Mhm

  53. Miracrown
    • one year ago
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    since you know those are equal, you can set them equal to solve for vo

  54. Jarp0120
    • one year ago
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    yay!!!! Vo=192

  55. Miracrown
    • one year ago
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    Hooray! lol

  56. Miracrown
    • one year ago
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    and that's in units of ft/sec rememeber

  57. Miracrown
    • one year ago
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    Now, can you plug that into the d equation and try to use it to maximize the height?

  58. Jarp0120
    • one year ago
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    yes thank you!!

  59. Miracrown
    • one year ago
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    yw :D

  60. Jarp0120
    • one year ago
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    salamat po!!

  61. Jarp0120
    • one year ago
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    yes thank you for your help!!

  62. Miracrown
    • one year ago
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    No worries. But hand on, It wants you to find the maximum height too doesn't? Alternately, you can maybe guess when the height is maxed If its the same height at t=2 and t=10 When do you think its at the top of its arc?

  63. Jarp0120
    • one year ago
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    wait

  64. Miracrown
    • one year ago
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    tyt

  65. Jarp0120
    • one year ago
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    6 sec? when v=0 right?

  66. Jarp0120
    • one year ago
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    :)

  67. Miracrown
    • one year ago
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    Yes! Yes! Yes! Yes!

  68. Miracrown
    • one year ago
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    But wait there dear

  69. Miracrown
    • one year ago
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    That'll be when its at the top you want to find the height then (so plug in to get s(6))

  70. Jarp0120
    • one year ago
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    when Vo=192?

  71. Miracrown
    • one year ago
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    Yupee. Vo is and always will be 192 now

  72. Jarp0120
    • one year ago
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    and then

  73. Miracrown
    • one year ago
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    You just want to find s(6)

  74. Jarp0120
    • one year ago
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    what will i do find s again? with t=6?

  75. Miracrown
    • one year ago
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    Yes

  76. Jarp0120
    • one year ago
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    576 ft

  77. Miracrown
    • one year ago
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    that will give you maximum height, and then youll be done

  78. Miracrown
    • one year ago
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    Yes, 576ft is the max height. You got it girl! :)

  79. Jarp0120
    • one year ago
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    thanks again but there are still many worded problem to be solved in my book hahah

  80. Miracrown
    • one year ago
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    Hahaha. #thestruggleisreal

  81. Jarp0120
    • one year ago
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    #calculusislove hahahah

  82. Miracrown
    • one year ago
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    Welp. You have a great night/day. Keep working hard and don't ever give up!

  83. Jarp0120
    • one year ago
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    i won't thank you

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