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This Projectile motion question seems flawed

  1. Sep 24, 2016 #1
    [moderator note: Thread moved from General Physics so no template shown]

    QUESITONWEIRD.png

    a) Considering X and Y velocities are independent of one another, simply getting the 300m it covered on the X axis and the total time it was in the air, 21.0s, it's easy to get an X velocity of 14.3 m/s to the right.

    b) I HAVE NO IDEA. It is thrown up, reaches an unknown height with an unknown time and comes back down. By the time it reaches the 150m mark, it already has a significant vertical velocity.
    The only thing we really have for this is the horizontal velocity and vertical acceleration.
     
    Last edited by a moderator: Sep 25, 2016
  2. jcsd
  3. Sep 24, 2016 #2

    A.T.

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    There is more given.
     
  4. Sep 24, 2016 #3
    For the parabolic path above 150m, I honestly don't know what else you have. There is more for the total path, but I don't know anything we can use from that to get the initial vertical velocity.
     
  5. Sep 24, 2016 #4

    A.T.

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    What info did you use for a) ?
     
  6. Sep 24, 2016 #5
    Total time of flight (parabolic path time + downward time), and total horizontal displacement...?
     
  7. Sep 24, 2016 #6
    Need to use vertical projectile motion equation.
     
  8. Sep 24, 2016 #7
    Why? Could you please explain the mathematical process?
     
  9. Sep 25, 2016 #8
    Been a while since I took physics, but your initial height is 150m, time is 21s, and you need to solve for Vo. There are only 3 variables in that equation. Use quadratic formula to solve for t and plug everything in.
     
  10. Sep 25, 2016 #9
    Sorry, don't solve for t. Solve for Vo. Duh me
     
  11. Sep 25, 2016 #10
    y=1/2*a*t^2 + Vo*t + yo

    Where:
    y=height(vertical distance traveled)
    t=time
    a=acceleration due to gravity
    Vo=initial velocity
    yo=initial height

    Solve equation for Vo and plug in your values.
     
  12. Sep 25, 2016 #11

    haruspex

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    Either y is the vertical displacement (distance travelled, but paying attention to signs) and y0 is redundant, or y0 is initial height and y is final height (again, making sure sign usage is consistent: up must be positive for all displacements, velocities and accelerations, or down must be positive for all).
     
  13. Sep 25, 2016 #12
    It doesn't matter which way is positive for you.

    They are just variables, call them a, b, c, etc. if you want.
    yo in this case would be your initial height, 150m in this case. You have to take into account the total flight time, which includes the 150m down from your initial height. Your ending y will be negative because you are throwing it off a cliff.
     
  14. Sep 25, 2016 #13

    haruspex

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    No, but I did not say it did. I wrote that you have to be consistent about which way is positive.
     
  15. Sep 25, 2016 #14
    Positive could be down or up, but yes it must be consistent for everything.
     
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