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More Free Fall!

  1. Nov 19, 2005 #1
    A hot air balloon is ascending straight up at a constant speed of 7.0 m/s. When the balloon is 12.0m above the ground, a gun fires a pellet straight up from the 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 the ground level are these places?
    For this problem, I need to find two distances, so I tried to find the two times for the distances first but I can't work it out. This is what I have:
    let v=velocity of balloon, t=time, V=velocity initial, a=acceleration
    then I factored it
    t=0, 4.68 or 4.7
    I'm missing one more time and distance or is this time also incorrect? I don't know how to solve this problem, plz help me out :redface:
  2. jcsd
  3. Nov 19, 2005 #2


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    you need to decide WHEN you're going to start timing ...
    either when the balloon leaves the ground,
    or when the bullet leaves the ground.

    I would start the timer when the pellet is fired.
    Where is the balloon when MY stopwatch reads zero?
  4. Nov 20, 2005 #3
    I redid the problem but I still can't get it because the two times are the same. What did I do wrong now?






  5. Nov 20, 2005 #4


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    1) why did you write vt - 12? When the bullet is fired the balloon is 12m over the ground.
    2) The initial speed of the bullet is 30 + 7 (which is the speed of the balloon) - unless your using some other referance system.
    3) Don't forget to add the bullets initial position to it's position equation.
  6. Nov 20, 2005 #5
    i don't understand why the speed of the balloon needs to be added to the initial speed of the bullet.
    Last edited: Nov 20, 2005
  7. Nov 20, 2005 #6


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    you should have 12[m] + 7[m/s] t = 30[m/s] t - ... ,
    which SHOULD result in 0 = -12[m] + (30 - 7)[m/s] t ...

    your writing would be a lot easier to read
    if you started sentences with the SUBJECT

    y(balloon) = 12[m] + v_i t
    y(pellet) =

    Don't try to write one very complicated sentence
    that includes all the important information in some hidden form!
    Write a few small statements, instead, so you won't get confused.
    (and you're allowed to use words to explain).
    Last edited: Nov 20, 2005
  8. Nov 21, 2005 #7
    12.0[m] + 7.0[m/s] t = 30.0[m/s] t - 0.5 (-9.81[m/s^2]) t^2
    0 = -4.91[m/s^2] t^2 + 23.0[m/s] - 12.0[m]


    product=59 sum=23.0

    the two numbers are 2.95 and 20.05

    (-4.91 t^2 + 2.95t) (20.05 t - 12.0)
    t (4.91 t + 2.95) (20.05 t - 12.0)


    4.91 t + 2.95 = 0
    4.91 t = -2.95
    t = 0.601

    20.05 t - 12.0 = 0
    20.0 t = 12.0
    t= 0.599

    now I substitute the times in for distance

    12.0[m] + 7.0[m/s] t
    12.0[m] + (7.0[m/s]) (0.601)
    12.0[m] + 4.207[m]
    = 16.207[m]

    12.0[m] + 7.0[m/s] t
    12.0[m] + (7.0[m/s]) (0.599)
    12.0[m] + 4.193[m]
    = 16.193[m]

    That's only a difference of 2cm (which makes me feel like it's incorrect) but is this correct. Plz let it be
  9. Nov 21, 2005 #8


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    This is correct.

    You should now just use the quadratic formula.

    you have,

    at² + bt - c = 0

    t = {-b +/- sqrt(b² - 4ac)}/2a

    You should get t = 0.6s and t = 4s (approx)
  10. Nov 21, 2005 #9
    I got the same answer (0.6 and 4) now, but I used a graphing calculator; I don't know how to get to t={-b+/- sqrt.... because I haven't learned how yet.
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