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Projectile Motion Diver Jump Question

  1. Sep 28, 2013 #1

    lvn

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    A diver springs upward from a board that is 3.60 m above the water. At the instant she contacts the water her speed is 13.1 m/s and her body makes an angle of 67.3 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

    Basically what I've tried is I found initial velocity component X by finding the final velocity X component:

    Vx = Vox

    Vx = 13.1cos67.3

    Vox = 5.06m/s


    Then I found the initial velocity Y component through the kinematic equation:

    Vy^2 = Voy^2 - 2a(y-yo)

    Voy = sqrt(171.61 - 2(-9.8)(-3.6))

    Voy = sqrt(171.61 - -19.6(-3.6))

    Voy = sqrt(171.61 - 70.56)

    Voy = sqrt(101.05)

    Voy = 10.05m/s

    Using pythagoras to find Vo:

    Vo = sqrt(10.05^2 + 5.06^2)

    Vo = sqrt(101.0025 + 25.6036)

    Vo = 11.25m/s

    Theta:

    Tan^-1 (Voy/Vox)

    = 96

    Answer was wrong - what am I doing wrong?
     
    Last edited: Sep 28, 2013
  2. jcsd
  3. Sep 28, 2013 #2

    lvn

    User Avatar

    Figured it out.
     
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