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Dropped from a hot air balloon

  1. Mar 6, 2009 #1
    1. A hot-air balloon is ascending at the rate of 11 m/s and is 53 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

    2. v = u + at
    s = ut + 1/2 a t^2
    v^2 = u^2 + 2as

    3. s = 53m
    u = 11 m/s
    v =
    a = -9.8 m/s^2
    t =

    So far I've tried v = u+at, but stoppped when i realised I had 2 unknowns.
    ( v = 11 + -9.8 t )

    I've also tried s = ut + 1/2 at^2 to no avail
    ( 53 = 11t + 1/2 (-9.8) t^2 )

    Also tried v^2 = u^2 + 2as but I got something like -1000 so it didn't make sense to me.

    I've hit a mental block, and it doesn't look like i's going to move any time soon, which isn't helpful since this thing is due in 10 hours. Sorry if I've put this in the wrong place or with the wrong title, because right now I'm not sure what it is, oher than a torture device. The answer will probably turn out to be easy to get, but right now it isn't working for me. Any and all help would be greatly appreciated.
  2. jcsd
  3. Mar 6, 2009 #2


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    Hi AlKaiser,

    I believe this equation will give you the time for the package to hit the ground.

    Your displacement is not quite right here. If the package falls from the balloon and hits the ground, the displacement is not 53 m; what would it be?
  4. Mar 6, 2009 #3
    Ok, so if it's not 53 metres, does that mean that the crate travels up before it travels down? Or is it -53m? I'm sorry, but I honestly am stumped on this one. I'm going to need a bigger hint because I really don't see how it isnt 53. (I'm not askin for the answer, just a bit more of a hint, as I'm not getting any closer atm)
  5. Mar 6, 2009 #4


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    Depends on where 0 is.

    If ground is 0 and you drop it from 53 then your Yo is 53. Otherwise if you drop it from 0, then it lands at - 53. Your choice, but once chosen stick to it like glue.
  6. Mar 6, 2009 #5


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    The displacement includes the direction, and so since the displacement is in the same direction as the acceleration (downward), it would need to have the same sign as the acceleration.

    (For these 1-dimensional problems, everything in the same direction has to have the same sign, and things in opposite directions have to have opposite signs.)

    That's certainly true of course, but the form of his equation had the displacement instead of the initial and final position.
  7. Mar 6, 2009 #6
    I think you'll have to take u as negative because baloon is travelling up.When it goes down, g is not -9.8m/s^2, but +9.8 m/s^2, because it is positively accelerating.
    Its a quadratic equation!
    Youll have to take the positive root.

    Now, you can calculate v.(v=u+at)

    Its simple, hope Im not wrong.

    Can someone say whether I am correct or not?
    Last edited: Mar 6, 2009
  8. Mar 6, 2009 #7


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    Right you are. I just never use S.

    Btw, good to see you back.
  9. Mar 6, 2009 #8
    Look at it this way; the balloon and its contents are ALL ascending at 11 m/s therefore, at the very instant that the package is released, it must slow from its 11 m/s ascent to its apex prior to being able to free-fall back to earth. You have this ascent time as well as the free-fall time from the apex as your total time.
  10. Mar 6, 2009 #9


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    Thanks! A few cruel turns in life had conspired to keep me from spending time at this wonderful place (I'm sure you can imagine how hard staying away from here would be). I've got a few more trips that will interfere in the near future, but I hope to be visiting here daily again very soon.
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