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Projectile Motion

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    y = y0 - 1/2gt^2 + voy*t

    3. The attempt at a solution
    Well there are two methods that I used for solving this, but I think only one is correct:

    First, the correct method:
    1) Find the time at the top of the parabola and use the distance from the hoop to determine the time.
    12 = 5 - 16t^2 + v0*t*(5/13)
    x = v0*(12/13)*t = 24
    t = .433s

    v0 is 60 ft/s

    2) Find the time at y = 3 (to hit the second slope)
    0 = 5-3 - 16t^2 + vo(5/13)*t
    t = 1.524s

    3) Find h
    x = vo(12/13)*t = h + 24 --> h = 60.4 ft

    I think everything I did is correct, however, my TA in lecture today said that the v0 should be between 50 ft/s - 60 ft/s and h should be between 40 ft and 47 ft. However, he also said that he didn't quite know how to get the correct solution.... (he was looking at the solutions manual)

    I get 60ft/s but my h is 60.4 ft however, if I use the 2nd method I get within the range, but isn't my method correct and the second method wrong?

    method 2:
    12 = 5 - 16t^2 + v0*t*(5/13)
    v = vo(5/13) - gt
    v = 0 since it's at the top of it's path
    v0 = gt * 13/5
    v = 55 ft/s ...
    t = .661 s
    But isn't that wrong because if I plug in .661 s into x = vo(12/13)*t I get ~33ft, whereas it should be 24 ft at the top of the path..

    Can anyone advise me on which method is correct?
  2. jcsd
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