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Really Need Help here with 2D Projectile Motion

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    A diver springs upward from a board that is 3m above the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 75 degrees with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

    Initial velocity = V1 , Final Velocity = V2

    Assuming Down as positive and up as negative
    Y direction X direction
    Dy = 3m. Dx = Unknown.
    Ay = 9.8 Ax = 0
    v2y = Sin 75 8.90 v2x = Cos 75 8.90
    v1 = unknown v1x = v2x ( since there is no accelaration ! )



    2. Relevant equations
    V2^2=V1^2 + 2ad



    3. The attempt at a solution

    Since V1x = v2x , We do not need to calculate V1x since its already 2.3 ( cos 75 8.90 )

    V2Y = ( Sin 75 8.90 ) = 8.60

    V2y^2= V1y^2 + 2ad
    (8.60) = V1Y^2 + (2x9.8x3)
    (8.60)^2-(2x9.8x3) = V1y^2
    3.89 = V1y^2

    Since 3.89 is positive, it should be pointing down. But when i make a vector with y point down and x pointing to the right, after using pythagoras theorem and tan , i get the wrong degree. However, if i assume down to be as negative and up to be as positive, then my direction is correct.
    I have always been told by my teachers that its upto me what i take as negative and positive. I want to know why im not being able to get the answer when i take down as positive and up as negative. either way i get my Y value as positive whether i take positive down or up , so how do i know which one is right.

    Help will really be appreciated. Thanks.
     
  2. jcsd
  3. Oct 17, 2011 #2
    Could someone delete this thread? I got the answer. Thanks!
     
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