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Angular relationship question.

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A model rocket is launched at 50m/s, 35 degrees above horizontal. What is the horizontal displacement when its velocity vector is at 25 degrees.


    2. Relevant equations
    Kinematic equations, the relationship between sin cos and tan.


    3. The attempt at a solution
    tan(25)= .4663. I interpret this as the y (sin) velocity being .4663 the x velocity. Vy = .4663Vx

    Vy = 50sin(25)-9.8t
    .4663Vx = 50sin(25)-9.8t
    Vx = (50sin25-9.8t)/.4663
    Vx also equals 50*cos25 therefore...
    50cos25 = (50sin25-9.8t)/.4663
    (.4663*50cos25-50sin25)/-9.8 = t
    t = 3.54E^-5.

    Something about this answer seems very very wrong, especially when plugged back in to the x displacement equation.
     
  2. jcsd
  3. Sep 9, 2009 #2

    ideasrule

    User Avatar
    Homework Helper

    Why 50? The rocket decelerated due to gravity.
     
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