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Odd projectile motion question.

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Question: "The time of flight of the lure through the air is 3.4 seconds. (g= 9.8m/s/s)
    Calculate the maximum height of the lure in its projectile motion."


    2. Relevant equations
    Not sure if I need to use this:
    d = (u)(t) + (1/2)(a)(t^2)


    3. The attempt at a solution

    I don't know how this really works...

    I'm guess to half the time (3.4/2). Thats the time when the object is at rest for a very short time (initial velocity would be 0). And use that equation on to solve the d.
     
  2. jcsd
  3. Sep 1, 2012 #2

    Borek

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    Staff: Mentor

    Once the object is at rest (top of the trajectory) what follows (on the vertical axis) is just a free fall, isn't it?
     
  4. Sep 1, 2012 #3

    CAF123

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    Gold Member

    You could start by finding the initial vertical component of velocity and using another kinematic relation to find the height. Or do as Borek suggested.
     
  5. Sep 1, 2012 #4
    So...

    d = (0)(1.6) + (1/2)(9.8)(1.6)^2

    = 12.544 m

    Is it right?
     
  6. Sep 1, 2012 #5
    The time of flight of the lure through the air is 3.4 seconds means the lure left the earth at t=0 and landed back to earth at 3.4sec.

    So vertically the final position is back to where it started.
    Y position is a horizontal line parallel to x-axis.
     
  7. Sep 1, 2012 #6

    Doc Al

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    Staff: Mentor

    Right idea, but 3.4/2 ≠ 1.6. :wink:
     
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