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Projectile Velocity

  1. Jun 4, 2007 #1
    A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.00 m/s and her body makes an angle of 81.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.
    magnitude m/s
    direction °

    Here's what I've tried to do: I tried to solve for the initial velocity by using the kinematics equation v^2=vo^2+2*a*x
    vo=1.174<<initial velocity

    Now I need to find the magnitude and the direction, I think you use tan-1 and pythagorean theorem but I don't really understand the component method very well. Could you please help me get on the right track for this problem. All help will greatly be appreciated.
  2. jcsd
  3. Jun 4, 2007 #2
    do you know the position function?
  4. Jun 4, 2007 #3
    no I haven't heard of the position function.
  5. Jun 4, 2007 #4
    do you know the kinematic equations?
  6. Jun 4, 2007 #5


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    How did you get the initial y velocity 1.174 m/s?
  7. Jun 4, 2007 #6
    Yes I know the kinematics equations x=vot+a/2at^2

    I got 1.74m/s for the initial velocity by using the fourth equation. I am not positive that I did it right. Not sure what to do next!
  8. Jun 4, 2007 #7
    in the y-direction ...

    v_{fy}^2 = v_{oy}^2 - 2g(Delta y)

    v_{oy} = sqrt[v_{fy}^2 + 2g(Delta y)]

    v_{oy} = sqrt[(9*sin(81))^2 + 2(9.8)(-3)] = approx 4.5 m/s

    velocity in the x-direction is a constant ...

    v_x = 9*cos(81) m/s

    you now have both x and y components of the initial velocity ... finish up.
  9. Jun 4, 2007 #8
    Thanks for all of your help!!
  10. Jun 5, 2007 #9


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    Yes, I thought that you did use the fourth equation, but I got a different answer. I was hoping that you would show more exactly how you susbtituted to get to it so that we can compare notes.
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