Constant Acceleration  Sandbag Dropped From a Rising Balloonby Nickg140143 Tags: acceleration, balloon, sandbag 

#1
Sep1311, 04:16 PM

P: 30

1. The problem statement, all variables and given/known data
A hot air ballonist rising verticlly at a constant velocity "v_{b}", releases a sandbag at the instant the sandbag is "h" meters above the ground. Given [v_{b},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 x_{0}=h? (starting position) not too sure about this one x_{1}=0 (ending position) v_{0}=v_{b} (starting velocity) v_{1}=? (ending velocity) a=g (acceleration) since its heading towards the ground? t= (time) time it hits the ground (when x_{1}=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
Sep1311, 04:39 PM

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#3
Sep1411, 01:23 AM

P: 30

Encouragement like that helps keep me going
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 x_{1}=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 v_{1} 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. 



#4
Sep1411, 05:39 AM

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Constant Acceleration  Sandbag Dropped From a Rising Balloon
There are a couple of things I need to point out.
1. Be careful when you apply the equation v^{2}=v_{0}^{2}+2aΔx. More transparently it should be written as v_{1}^{2}v_{0}^{2}=2a(x_{1}x_{0}) in which subscripts "1" and "0" refer to different points in space, not time. Then if you identify quantities as you have, you get v_{1}^{2}v_{b}^{2}=2(g)(0h), from which you get v_{1}^{2}=v_{b}^{2}+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 (x_{1}x_{0}) 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 = x_{0}+v_{0}t+(1/2)at^{2}. It gives you position (not displacement) as a function of time. If you set x_{0} = h, it gives the position of the sandbag above ground at any time t. There is a specific time t_{f} at which the sandbag hits the ground. This means that if you replace "any time t" with t_{f} in the above equation, the position of the sandbag above ground will be ...? Note that you have a quadratic to solve for t_{f}. 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. 



#5
Sep1411, 11:05 AM

P: 30

All right, I did a little reworking to see if I could fix those mistakes. Looking at the work I did to solve for v_{1} 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? 



#6
Sep1411, 12:02 PM

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Mistakes
1. Your velocity solution (in red) has a mistake. Somehow you converted v_{0}^{2} in the original equation to h^{2}. It should be v_{b}^{2}. 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 v_{b}t 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 v_{b} 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. 



#7
Sep1411, 01:03 PM

P: 30

I was messing with those calculations again while I was in class just now, and noticed those (rather careless) mistakes
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. 



#8
Sep1411, 02:58 PM

P: 30

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 v_{1}^{2}v_{0}^{2}=2a(x_{1}x_{0}), since it's velocity as a function of displacement, I figured it might be possible to set x_{0} in the equation to "h" and v_{1} to 0, then solve for x_{1}. Am I correct in my reasoning? 



#9
Sep1411, 03:13 PM

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Everything looks just fine. You should be proud of yourself because you did well with minimum guidance.




#10
Sep1411, 03:37 PM

P: 30

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. 



#11
Sep1411, 04:27 PM

P: 159

Thanks everyone here! Me and Nick must be in the same class because this was a huge help to me as well!




#12
Sep1411, 05:31 PM

P: 30

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. 


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