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Projetile Motion

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A soccer ball is kicked from the ground at an angle theta above the horizontal. Show that the equation h = 0.25R tan theta represents the maximum height of the ball, where h is the height and R is the range.


    2. Relevant equations

    t= dx/V1x
    dy = v1y(t) + .5a(t)^2

    3. The attempt at a solution

    Vert - dy = 0, V1 = vsintheta, a = g, t=?
    Horizontal - dx = R, V1 = vcostheta, t=R/vcostheta

    dy = v1y(t) + .5a(t)^2
    0 = vsintheta(R)/vCostheta + 1/2g(R/vcostheta)^2
    sintheta(R)/Costheta = -1/2g(R/vcostheta)^2

    Help plz
     
  2. jcsd
  3. Sep 13, 2007 #2
    anyone?
     
  4. Sep 13, 2007 #3

    learningphysics

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    Looks good... except you should have used -(1/2)gt^2...

    Get another equation for maximum height...
     
  5. Sep 13, 2007 #4

    Doc Al

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    Realize that that's the time for the complete trajectory. How long does it take to reach the maximum height?
     
  6. Sep 13, 2007 #5

    learningphysics

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    One simplification I wanted to point out:

    0 = vsin(theta)t - (1/2)gt^2 (where t is the time to reach R)

    can be simplified to

    0 = vsin(theta) - (1/2)gt, where t = R/(vcos(theta))
     
  7. Sep 13, 2007 #6
    0.5R?

    I am completely loss and dont know where to go from the step i took it up to
     
    Last edited: Sep 13, 2007
  8. Sep 13, 2007 #7

    Doc Al

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    That's a distance, not a time. :wink:
     
  9. Sep 13, 2007 #8
    t= 0.5(R/vcostheta)?
     
  10. Sep 13, 2007 #9

    Doc Al

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    Good.
     
  11. Sep 13, 2007 #10
    so where does this take me? here?

    0 = sin(theta) - (1/2)(0.5(R/costheta))
    (1/2)(0.5(R/costheta)) = sin(theta)
     
    Last edited: Sep 13, 2007
  12. Sep 13, 2007 #11

    Doc Al

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    No.

    Keep it simple. You know how long it takes to reach maximum height. What's the average (vertical) speed as it rises?
     
  13. Sep 13, 2007 #12
    V1 = vsintheta?
     
  14. Sep 13, 2007 #13

    Doc Al

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    That's the vertical component of the initial velocity.
     
  15. Sep 13, 2007 #14
    um no clue =(
     
  16. Sep 13, 2007 #15

    Doc Al

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    What's the vertical component of the velocity when it reaches maximum height?
     
  17. Sep 13, 2007 #16
    (2d /0.5(R/vcostheta)) - vsintheta = V2?
     
  18. Sep 13, 2007 #17

    Doc Al

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    What if you threw a ball straight up? What would its speed be at its highest point?
     
  19. Sep 13, 2007 #18
    Zero
     
  20. Sep 13, 2007 #19

    Doc Al

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    Yes! Now apply that fact to your problem. You know the initial (vertical) speed and the final speed: so what's the average (vertical) speed as the projectile rises?
     
  21. Sep 13, 2007 #20
    vsintheta/2
     
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