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Resultant velocity

  1. Jan 26, 2005 #1
    Is the result of a horizontal velocity and a vertical velocity caluculated by a vertor diagram whereby the resultant velocity is at 45 degrees from the horizontal?

    If not, is is something to do with angles, please explain!
     
  2. jcsd
  3. Jan 26, 2005 #2

    dextercioby

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    Nope,45° will be opnly in the particular case in which the velocities will be equal in modulus...

    Apply the "paralelogramme rule"...Add the vectors and use the pythagorean theorem and the (circular) trigonometrical functions...

    Daniel.
     
  4. Jan 26, 2005 #3
    Ok, thanks for the reply, but this doesn't help me to much.

    The veloicties are directly along the vertical and horizontal axis.

    Why is the paralleogram rule needed?
    I get why you use pythogorean theorem(i think), and do you use the trig functions to calculate the eultant velocties angle from the vertical plane?

    So you use pythagorean theorem to find the magnitude of the resultant velocity.

    vertical V squared + horizontal V squared = resultant V squared

    And the angle form the vertical is calculated by using the tan rule of opp over adj.(The values of the opp and adj being the horizontal and vertical velocities).
     
    Last edited: Jan 26, 2005
  5. Jan 26, 2005 #4

    BobG

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    Exactly right.

    You're still using the parallelogram rule. It's just in this case, with the vectors being at right angles to each other, you happen to have a special case where your parellelogram happens to be a rectangle (this is the easiest case of the parellogram rule).
     
  6. Jan 27, 2005 #5
    I see what you mean now. Since the velocities are both on the horizontal and vertical planes, and drawing in the other lines would give you the parallelogram, or in this case, a rectangle.
     
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