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Solving for velocity

  1. Nov 29, 2012 #1
    1. The problem statement, all variables and given/known data
    A boy jumped off of a cliff into the water at a 26° angle. He traveled 23 meters and fell 12. He had 4.8 seconds of travel time. What was the velocity needed to do this?


    2. Relevant equations
    X=Vx + t
    Y=Vyt + 1/2at^2


    3. The attempt at a solution
    I wish I could, but I've only used 45° problems like this.
     
  2. jcsd
  3. Nov 29, 2012 #2

    haruspex

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    Is the angle measured from vertically up, upwards from horizontal, downwards from horizontal, or from vertically downwards? Assuming it's θ above horizontal, if the take-off speed is V, what would Vx and Vy be?
    Regarding the equations you quote:
    X=Vx + t
    That should be X=Vx t
    Y=Vyt + at2/2
    Need to be careful with the signs. First, define whether up or down is your positive direction, then use that consistently for distance, speed and acceleration.
     
  4. Nov 29, 2012 #3
    26 degrees upward from horizontal
     
  5. Nov 29, 2012 #4

    haruspex

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    So what would Vx and Vy be? (Preferably expressed in terms of an arbitrary angle θ, rather than specifically 26o.)
     
  6. Nov 30, 2012 #5
    distance = Vx . t where t = 4.8sec and distance in x-direction = 23 meters

    so Vx = 23 meters / 4.8sec = 4.79 m/s

    finally V = Vx / sin23 = 12.26 m/s assuming angle measured from verticle
     
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