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Basic question about speed

  1. Jan 31, 2009 #1
    alright, i know this is really basic stuff, so please bear with me.

    let's say i have a box or something that is sliding horizontally at 2m/s across a table. all of a sudden it falls 0.5m to the ground. what's the speed of the box right before it hits the floor?

    okay, so it's pretty easy to see that 0.5m will result in a vertical velocity component of sqrt(10) m/s. my question is, to get the speed, do i just add them up in quadrature? i.e.,

    speed = sqrt(2^2 + sqrt(10)^2) = sqrt(14)?
  2. jcsd
  3. Jan 31, 2009 #2
    umm I don't think so. I think all you need is the kinematics equation
    (final velocity)^2 = (initial velocity^2 + 2(acceleration)(distance)

    the distance is 0.5, acceleration due to gravity is 9.8 and the inital velocity is 0 m/s because the box slides off the table with 2 m/s horizontal velocity but no vertical velocity

    so (final velocity) = sqrt ( 2 x 9.8 x 0.5)

    does that make sense?
  4. Jan 31, 2009 #3
    i thought speed was a scalar. doesn't that mean it's the magnitude of the resultant vector of the two components?
  5. Jan 31, 2009 #4


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    I think you are right, syang9. NeedLottaHelp didn't realize you already had the vertical part done. Maybe use 9.81 instead of 10, though.
  6. Jan 31, 2009 #5
    so the answer is sqrt(14)?
  7. Jan 31, 2009 #6


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    Yes, if g = 10 in your part of the world.
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