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General Launch Angle (Vectors) HELP

  1. Nov 15, 2007 #1
    General Launch Angle (Vectors) PLZ HELP URGENT

    1. The problem statement, all variables and given/known data

    In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall witha speed of 13m/s at an angle of 24 degrees above the horizontal. a) how long does it take for the ball to reach the wall if it is 4.2m away? b) how high is the ball when it hits the wall? c) what are the magnitude and direction of the ball's velocity when it strikes the wall?

    2. Relevant equations

    magnitude - vector A is A = (square root Ax^2 + Ay^2)
    direction- vector A is theta = inverse tangent (Ay/Ax), theta is measured relative to x axis.

    3. The attempt at a solution

    I have already done a) .35 seconds and b) 1.36 meters... for c) i got the magnitude which is 12 m/s but i am clueless on how to get direction...somebody please help i'm lost and i have tried my best not to mention i have a quiz on this stuff in four hours...
  2. jcsd
  3. Nov 15, 2007 #2

    Shooting Star

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    You have already written down how to to find the direction of a vector in terms of Ay and Ax.

    Find vx and vy at the wall.
  4. Nov 15, 2007 #3
    how might i go about doing that please?
  5. Nov 15, 2007 #4

    Shooting Star

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    Calculate vx and vy separately. vx remains constant beacuse there's no force in the horizontal direction. vy changes due to gravity acting down ward. So,
    vy_final^2 = vy_initial^2 -2gh.
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