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One Dimensional Hot Air Balloon Kinematics

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A hot-air balloon has just lifted off and is rising at the constant rate of 2.3 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12 m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?


    2. Relevant equations

    One-Dimensional Kinematics


    3. The attempt at a solution

    First I tried setting the distances of the balloon and the camera equal to each other,

    camera: 12t+.5*-9.81t^2
    balloon: 2.3t+2.5

    12t+.5*-9.81t^2=2.3t+2.5

    then I rearranged the equation

    12t-4.905t^2-2.3t=2.5
    9.7t-4.905t^2-2.5=0
    -4.905t^2+9.7t-2.5=0

    Next, I plugged the equation into the quadratic formula.

    t= -9.7+-sqrt 9.7^2-4.905*-2.5/2*-4.905
    t= -9.7+-sqrt 94.09+19.62*-2.5/-9.81
    t= -9.7+-sqrt 94.09-49.05/-9.81
    t= -9.7+-sqrt 45.04/9.81
    t= -9.7+-6.71/9.81
    t= .305, 1.67

    next, I plugged the t value into the formula x= Xi + Vi*t + .5at^2
    x= 0 + 12*.305 + .5*-9.81*.305^2
    x= 3.66-4.905*.093
    x= 3.66 -.456
    x= 3.2m

    I entered this into the website we turn our homework into, but it said the answer was incorrect, I'm not really sure what I did wrong, all my math seems correct.
     
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 13, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Flippit! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Looks ok to me. :confused:

    (except I would have used the 2.3t+2.5 equation at the end, instead of the quadratic :wink:)
     
  4. Sep 13, 2009 #3
    Alright, I guess I'll just have to talk to my teacher about it. Thanks for looking it over!
     
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