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Kinematics of the human canon fly zaccini

  1. Apr 17, 2008 #1
    as a human canon fly zaccini flied a distance of 53 meter in 1940 ( still record )
    his velocity as he leave the cannon is 24,5 m/s in a angle a with thw horizontalplane

    a) what is the angle ? ( 31,2 )
    b) find the highest point during the record ( 8,04m)

    can you set up theese tasks ?
     
  2. jcsd
  3. Apr 18, 2008 #2
    Do anyone know the formula ?
     
  4. Apr 18, 2008 #3
  5. Apr 19, 2008 #4
    x = v0*cos(a)*t
    y = v0*sin(a)*t - 1/2*gt^2
    since when x = 53 m, y = 0
    v0*sin(a)*t - 1/2*gt^2 = 0
    so t = 2v0*sin(a)/g
    then x = v0*cos(a)*2v0*sin(a)/g = v0^2*sin(2a)/g
    sin(2a) = g*x/v0^2
    plug known values and get a = 30 degrees (not 31.2 (?))
    then we can find t.
    highest: find y when t' = t/2
     
  6. Apr 19, 2008 #5
    Thanks for answers
    butI dont get it :(
    g*x/vo^2 I get 0,88 ? ( 9,8*53/24,2^2)

    And the reason for 30 degress was that I did put in wrong velocity It should be 24,2m/s
     
  7. Apr 19, 2008 #6
    yes, but don't put 0.88 in your calculator. Just use calculator for 1/2 arcsin(g*x/vo^2) = 31.2
     
  8. Apr 19, 2008 #7
    Yes Thanks =) I got it now=) Found a formula on the link kamerling showed, I did not use arcsin just sin.
     
    Last edited: Apr 19, 2008
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