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Homework Help: Constant Acceleration - Sandbag Dropped From a Rising Balloon

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    A hot air ballonist rising verticlly at a constant velocity "vb", releases a sandbag at the instant the sandbag is "h" meters above the ground. Given [vb,h] Determine:
    a. The time for the sandbag to hit the ground
    b. The speed of the sandbag just before hitting the ground.
    c. The maximum height the sandbag attains above the earth.

    2. Relevant equations
    v=v0+at (velocity as a function of time)
    x=x0+v0t+(1/2)at2 (displacement as a function of time)
    v2=v20+2aΔx (velocity as a function of displacement)

    3. The attempt at a solution
    I believe that I'm having trouble setting up my problem correctly. Here are the givens:

    If the coordinate system is set so that the ground is 0, and the sandbags position after being let go is h

    x0=h? (starting position) not too sure about this one
    x1=0 (ending position)
    v0=vb (starting velocity)
    v1=? (ending velocity)
    a=-g (acceleration) since its heading towards the ground?
    t= (time) time it hits the ground (when x1=0)

    So before I go any further, does anyone see any problems with this?
    I'm aware that the sandbag will travel just a bit farther up after being dropped from the
    balloon, and that its velocity should be 0 at the peak of its height (right?).

    However, this leads me to ponder whether or not the starting position should be adjusted
    or stay as h.

    I've been stuck on this problem in my homework for 3 days, and there are 7 more problems after it 0__0. whether that homework becomes finished now is irrelevant, I feel my time is best spent understanding how to pull this off.
  2. jcsd
  3. Sep 13, 2011 #2


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    Everything is correctly identified.

    If the sandbag is released at a height of, say, 120 m, does the fact that it falls to ground affect the fact that it was at a height of 120 m when it was released?

    I agree with you and I applaud your attitude. I wish more students were like you. Good luck! :approve:
  4. Sep 14, 2011 #3
    Encouragement like that helps keep me going:biggrin:

    It's good to hear that I was able to at least set my givens correctly, onto the meat of the problem.

    My first thought is too see if I could at least find the velocity when at the ground, that is, when x1=0. So looking at the formulas I have, I thought that this would come in handy.
    v2=v20+2aΔx (velocity as a function of displacement)

    Please take a look at the image attached to this post to see what I did to calculate for v1 and my attempt to find t
    (I'm not very good with displaying my math work in the forums)
    As you'll probably notice, I'm not too sure about my math on this one.

    It looks like a quadratic, but I'm not necessarily sure how to go about solving for t. I'd like to know if I'm either approaching this correctly but having a little math difficulties, or if I should take my thinking in another direction.

    Attached Files:

  5. Sep 14, 2011 #4


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    There are a couple of things I need to point out.
    1. Be careful when you apply the equation v2=v02+2aΔx. More transparently it should be written as v12-v02=2a(x1-x0) in which subscripts "1" and "0" refer to different points in space, not time. Then if you identify quantities as you have, you get v12-vb2=2(-g)(0-h), from which you get v12=vb2+2gh. Do you see the difference between your expression and this one? Correctly written, Δx = -h because even though the initial height is positive, the displacement is negative. That is why I advocate not using Δx but (x1-x0) instead. Then you match velocity at one point with the corresponding position at the same point.

    2. Since it is time you are looking for, why not use the kinematic equation that involves time? Look at x = x0+v0t+(1/2)at2. It gives you position (not displacement) as a function of time. If you set x0 = h, it gives the position of the sandbag above ground at any time t. There is a specific time tf at which the sandbag hits the ground. This means that if you replace "any time t" with tf in the above equation, the position of the sandbag above ground will be ...? Note that you have a quadratic to solve for tf. Once you do that, you can go back to the kinematic equations to find the rest.

    Your approach to answer the question by finding the final velocity first is correct. You can do it that way. However, since the problem asks you to find the time of flight first, you might just as well do that.
  6. Sep 14, 2011 #5
    All right, I did a little reworking to see if I could fix those mistakes. Looking at the work I did to solve for v1 and t, it looks more coherent than my last attempts, does the work look alright? (work is in the attached image)

    I haven't dealt with the quadratic formula in a while, so do please let me know if I'm misunderstanding something in its application to equation for time. Also, what is the significance of getting two possible answers for time?

    Attached Files:

  7. Sep 14, 2011 #6


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    1. Your velocity solution (in red) has a mistake. Somehow you converted v02 in the original equation to h2. It should be vb2.
    2. In the position as a function of time equation, when you eliminated the 2 in the denominator, you multiplied h by 2 (correct) but you did not multiply vbt by 2 (incorrect).

    The second (negative) root that one normally discards signifies an earlier time than the actual release of the sandbag. It is the time required for the sandbag to reach height h with upward velocity vb after being released at ground level. In other words, there are two possible times when the sandbag is at zero height and the negative time is the earlier possibility.
    Last edited: Sep 14, 2011
  8. Sep 14, 2011 #7
    I was messing with those calculations again while I was in class just now, and noticed those (rather careless) mistakes :uhh:

    I'll finish this work up and post it when I can for checking, but it looks like its coming together now.

    Thanks for all the info so far kuruman, its been very helpful.
  9. Sep 14, 2011 #8
    Alright, worked out the equations again, this time I wrote out as many of the steps as possible, to help avoid any algebraic errors.

    Now if that seems to be correct, the last thing I need is the max height that the sandbag attains. My approach to this is to use the equation v12-v02=2a(x1-x0), since it's velocity as a function of displacement, I figured it might be possible to set x0 in the equation to "h" and v1 to 0, then solve for x1.

    Am I correct in my reasoning?

    Attached Files:

  10. Sep 14, 2011 #9


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    Everything looks just fine. You should be proud of yourself because you did well with minimum guidance. :smile:
  11. Sep 14, 2011 #10
    One question slain, I am pleased.

    Thanks for all the help Kuruman, the advice you've provided will be of great use to me. I'm sure you'll be seeing various other questions of mine in the forums within the near future. :biggrin:
  12. Sep 14, 2011 #11
    Thanks everyone here! Me and Nick must be in the same class because this was a huge help to me as well!
  13. Sep 14, 2011 #12

    I'm glad someone else was also able to get some help from our posts. This is the first time I've used this site, and I can already tell you thanks to people like Kuruman, I may just yet survive this class. :biggrin:
    Last edited: Sep 14, 2011
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