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Finding speed when fish jumps out of water

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Salmon, swimming up the Fraser river to their spawning grounds, leap over all sorts of obstacles. The unofficial salmon-altitude record is an amazing 3.44 m height jump. Assuming the fish took off at 45.0o, what was its speed on emerging from the water? Ignore friction.


    2. Relevant equations
    ?


    3. The attempt at a solution
    Vxo= Vo Cos (45)
    Vyo= Vo Sin (45)

    Time for fish to fall the 3.44 m = time for fish to jump 3.44 m

    Yv=0=3.44 - 1/2 (9.8) t^2
    -3.44=-4.9*t^2
    -3.44/-4.9= t^2
    .7020=t^2
    t= .8379 s

    final velocity = 3.44/ .8379
    =4.106 m/s

    Vyo=4.106
    4.106/sin(45)=Vo
    Vo= 5.807

    Vxo= 5.807 Cos(45)= 4.106

    it told me that it was the wrong answer. please help!!!
     
  2. jcsd
  3. Sep 23, 2008 #2
    The final velocity is *not* d/t. Remember the velocity varies over this distance.

    You say "Time for fish to fall the 3.44 m = time for fish to jump 3.44 m". Are you sure? I don't see why you need to bother considering the time for fish to fall. Can you do everything just considering the upward motion? Hint:

    Uy - vertical initial velocity as fish emerges from water, then:

    Uy = gt

    3.44 = Uyt - 1/2gt2
     
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