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Acceleration word problem help :/

  1. Sep 17, 2006 #1
    acceleration/velocity word problem help!

    The problem is:
    A hot air balloon is ascending straight up at a constant speed of 7.0 m/s. when the balloon is 12.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places?

    I've been stuck on this problem for hours.
    I started out using the equation d=Vi*t+1/2*a*t^2 for both the balloon and the pellet, then set them equal
    but I run into problems of where to include the 12.0m at.
    Ugh...where do I start?
    Last edited: Sep 17, 2006
  2. jcsd
  3. Sep 17, 2006 #2


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    You should use the full equation;

    [tex]d = d_{i} + v_{i}t + \frac{1}{2}at^{2}[/tex]

    Where di is the intial position. Set the displacements equal to each other and solve for t. Does that make sense?
  4. Sep 17, 2006 #3
    we've never learned that equation :confused:

    But I had the same idea, I just am confused about the balloon's displacement
    Would it be [tex]12m+7.0m/s*t[/tex]??
    Last edited: Sep 17, 2006
  5. Sep 17, 2006 #4
    The final equation I end up getting is 0=-4.9m/s^2*t^2-7.0m/s*t-12m
    When I solve that using the quadratic formula I get the squareroot of a negative number.
    What am I doing wrong?
    Last edited: Sep 17, 2006
  6. Sep 18, 2006 #5


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    That is spot on, so you should have;

    [tex]7t + 12 = 30t - 4.905t^{2}[/tex]

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