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Physics again

  1. Jan 30, 2008 #1
    [SOLVED] Physics again

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

    A small ball rolls horizontally off the edge of a tabletop that is 1.19 m high. It strikes the floor at a point 1.78 m horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

    2. Relevant equations
    x=vcos(x)t
    y=vsin(x)t- .5gt^2


    3. The attempt at a solution

    I believe those are the right equations that I need, but I have two unknowns in the equation, velocity and time.
     
  2. jcsd
  3. Jan 30, 2008 #2
    The only other equation that I could maybe see using is y=(tan(*)x - (gx^2)/2(vcos(*)^2)
     
  4. Jan 30, 2008 #3
    If the ball were simply dropped from a height of 1.19 m, how long would it take to fall?
     
  5. Jan 30, 2008 #4
    This may sound stupid, but I have no idea....
     
  6. Jan 30, 2008 #5

    rock.freak667

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    Homework Helper

    [tex]y=ut+\frac{1}{2}gt^2[/tex]
    (u=initial velocity)

    Vertically...what is the height it falls from? Is there any initial vertical velocity?
     
  7. Jan 30, 2008 #6
    Hmmm since there is no initial velocity, then the "ut" in essence cancels out to 0, thus, solving for t I got, .493 seconds.
     
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