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Projectiles launched at an angle.

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data

    In 1991, Doug Danger rode a motorcycle to jump a horizontal distance of 76.5m. Find the maximm height of the jump if his angle with respect to the ground was 12.0?.

    2. Relevant equations

    Initial Velocity= ?g?x/2(sin ?)(cos ?)
    ?t= ?x/vi(cos ?)
    ?y= vi(sin ?)?t+1/2g(?t)^2
    3. The attempt at a solution

    vi= ?(9.81m/s/s)(76.5m)/2(sin 12?)(cos 12?)
    vi= 43.3 m/s

    ?t= 76.5m/(43.3m/s)(cos 12?)
    ?t= 1.8s

    ?y= (43.3m/s)(sin 12?)(1.8s)+.5(-9.81m/s/s)(1.8s)^2

    Not sure if this is right...
    Last edited: Oct 16, 2007
  2. jcsd
  3. Oct 16, 2007 #2
    I got V = 60.74 m/s and t = 2.575 s. Your answer of 43.3 m/s and 1.8 seconds will produce 76.23 meter horizontal distance.
    I used Vcos(12)*t = 76.5 and 0 = Vsin(12) - gt to get t = Vsin(12)/g... so t = 0.0212V, multiplied by 2, t = 0.424V, then, V = 13.58 m/s, t = 5.758 s from here you may use

    H = Vcos(12)t - 0.5gt^2
  4. Oct 16, 2007 #3
    Thanks for the help.
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