Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook