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Homework Help: Kinematics in one dimension - hot air balloon and bullet

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A hot air balloon is ascending straight up at a constant speed of 8.10 m/s. When the balloon is 13.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 28.0 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? Enter the answers in the ascending order.

    2. Relevant equations

    Kinematic equations:
    V = Vo + at
    X - Xo = Vot + .5at2
    v2 = vo2 + 2a(X - Xo)
    X - Xo = .5(Vo + V)t

    3. The attempt at a solution

    I tried using the X - xo = Vot + .5at^2 equation for both the balloon and the bullet to get two variables, delta x and t, and then solving for each. But I've tried it several times and I keep getting funky answers (like negative values for t) and I don't know what I'm doing wrong.
  2. jcsd
  3. Sep 14, 2010 #2


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    Welcome to PF,

    Can you post what you tried? What was your equation for the height vs time of the balloon? And of the bullet?

    What do you have to do with these equations in order to figure out what the problem is asking for?
  4. Sep 14, 2010 #3
    Sure :)

    So in my latest attempt --

    V = 8.10 m/s
    a = 0 m/s^2

    x = Vot + .5at^2

    (I am assuming there is no acceleration because velocity is constant, and I am putting in the velocity of the balloon as initial velocity, not sure if this is a correct assumption or not...)

    Vo = 28 m/s
    a = -9.8 m/s^2 (acceleration due to gravity, in this problem I am assuming up is positive so acceleration is negative)

    x = Vot + .5at^2
    x=(28)t - 4.9t^2

    Setting them equal to each other...

    8.1t = 28t - 4.9t^2
    4.9t^2 - 19.9t = 0
    t (4.9 t - 19.9) = 0 giving one solution t=0
    4.9t - 19.9 = 0
    4.9 t = 19.9
    t = 4.06 s

    Then I subbed into the balloon equation..

    x = 8.1 t
    x = (8.1)(4.06)
    x = 32.9
    and x = 0 when t=0, but adding the 13 m of the balloon height giving
    x = 13, x = 32.9

    So I know it's wrong and I probably made a bunch of different mistakes, but I am feeling reallyyyy lost

  5. Sep 14, 2010 #4


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    Yes, all of your assumptions are, of course, correct. However, it would probably be easiest if you chose t = 0 to be the moment at which the bullet is fired. Therefore, the "initial" position of the balloon is not 0 m, but rather 13.0 m. In other words, we have a non-zero x0 term in the equation:

    x = x0 + v0t

    where x0 = 13.0 m.

    This looks okay to me as well. Try solving it again, this time taking into account the initial height of the balloon, which you forgot.
  6. Sep 14, 2010 #5
    Worked perfectly -thank you very much!
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