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Projectile motion (I think)

  1. May 31, 2007 #1
    Q: A small ball rolls horizontally off the edge of a tabletop that is 1.23 m high. It strikes the floor at a point 1.96 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?

    So this is what I did, i treated the vertical and horizontal motions seperately. And for (a) I solved for the time by using the constant acceleration equation, with y-yo=-1.23, a=-9.8 m/s^2 and Voy=0. Using those values I got t= .50s. I then analyzed the horizontal motion, but I'm not understanding how to get the speed at the instant it leaves the table since at t=.50s would be the speed at which it hits the ground, how do I get the time at the instant it leaves the table in order to use the constant acceleration formula? Any help at all is appreciated, thanks
     
  2. jcsd
  3. May 31, 2007 #2

    Doc Al

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    Staff: Mentor

    Good.
    How does the horizontal component of velocity change with time?
     
  4. May 31, 2007 #3
    the horizontal component of velocity is constant.
     
  5. May 31, 2007 #4
    Does that mean that I can take the velocity at any time t?
     
  6. May 31, 2007 #5
    Or rather time t=.5s using x-x0=1.96 m
     
  7. May 31, 2007 #6
    Yep, that was right!!! Thanks alot Doc Al.
     
  8. May 31, 2007 #7

    Doc Al

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    Right!

    I'm not sure what you mean, since it's a constant. How do you propose to find the horizontal velocity?
     
  9. May 31, 2007 #8

    Doc Al

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    Right. That will give you the horizontal velocity period, not just at t=.5s.
     
  10. May 31, 2007 #9
    Ok, I see what you mean now, it just made since to use t=.5s because we had a horziontal distance for that value, Thanks again
     
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