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Motion in 2 dimension

  1. Sep 21, 2007 #1
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    A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8m/s at an angle of 20 below the horizontal. It strikes the ground 3.00s later. The horizontal distance at which the balls strike the ground from teh base of teh building and the height from which the ball was thrown are respectively given by:

    Xf = Xi + Vxi*t

    Change of X = Vxi*T

    Xf = 8cos(20)(3)

    Xf = 22.5 m

    and Yi

    Y = Vyi - 1/2g*t^2

    0 = Yi + 8sin(20) - 1/2(9.81)(3)^2

    Yi= 36m

    are my answers right?
     
  2. jcsd
  3. Sep 21, 2007 #2

    Gokul43201

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    This is correct.

    There are 2 errors in the bolded lines:
    1. Check the equation - something's missing;
    2. Check the signs of all terms - which way is the positive y-axis pointing?
     
  4. Sep 21, 2007 #3
    hmmm...

    Yf = Yi + Vyi - 1/2gt^2

    0 = Yi + 8sin(20) - 1/2(9.81)(3)^2

    -Yi = -36

    yi = 36...I can't think of anything mistake here but my book is saying t hat I got the wrong answer...
     
  5. Sep 21, 2007 #4

    learningphysics

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    As Gokul mentioned there are 2 things wrong with your formula... first off, basically you've got the formula wrong... check what the formula is... second a sign is wrong...
     
  6. Sep 22, 2007 #5
    Yf = Yi + Vyi - 1/2gt^2

    0 = Yi + -8sin(20)(3) - 1/2(9.81)(3)^2

    -Yi = -52.30

    yi = 52.30 m?

    now?
     
  7. Sep 22, 2007 #6

    learningphysics

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    Yes. Looks right.
     
  8. Sep 22, 2007 #7

    Gokul43201

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    Although you've plugged in correctly, the first line should read: Yf = Yi + Vyit - 1/2gt^2
     
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