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Homework Help: Distance Between 2 Objects

  1. Nov 7, 2007 #1
    A hot air balloon is rising at 5.0 m/s when a bag is released. Five seconds later, how far below the hot air balloon is the sand bag. Answer: 122.5 m.

    So I use the equation (1/2)(a)(t)^2 to figure out the distance the bag has traveled. I use the equation -Vt to figure out how far the balloon has risen. I end up getting 97.5 m when I combine the 2 but when I use just (1/2)(a)(t)^2, I get the correct answer.. Can someone please explain to me what I'm doing wrong?
    Last edited: Nov 7, 2007
  2. jcsd
  3. Nov 7, 2007 #2
    did you take into account that when the sandbag begins to fall it has an upward velocity of 5m/s?
  4. Nov 7, 2007 #3
    Wow! I can't believe I forgot about that. Thanks a lot.
  5. Nov 7, 2007 #4

    Chi Meson

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    The actual equation is [tex]d=v_o t + 1/2at^2[/tex]
  6. Nov 7, 2007 #5
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