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Homework Help: Physics Falling Bodies Question

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A person in a hot air balloon that is rising accidentally drops a can of paint from the balloon. The balloon is moving up at a constant velocity of 4m/s, the distance to the ground is at the moment 4m. Find the time it takes for the can to drop to the ground.

    2. Relevant equations
    v2^2=v1^2 + 2ad
    a = (v2-v1)/t

    3. The attempt at a solution
    To try to solve the equation, I first started off by using the first formula and rearranging to solve for (t) while leaving v1 as 0, and using 9.8m/s (grav) for (a). However, I quickly realized that v1 is actually 4m/s, so I quickly changed my solution.

    After I solved for (t) again with the new value for v1, I got a really small number. This seemed wrong, so I tried instead the second equation to solve for v2, then I input that into equation 3, but it was also a small number, I got 0.58. My friend got 1.4s instead, and I am pretty sure he is right.

    Can anyone point me in the right direction? I didn't review physics 11 yet, and I am struggling to recall the procedure for the initial motion questions. Thanks for any help.
  2. jcsd
  3. Sep 13, 2007 #2


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    Yes, use that first formula... careful about signs and directions...
  4. Sep 13, 2007 #3
    When I rearranged the formula to solve for (t), I got:

    2d/(v1*a) = t^3

    Assuming that I the (ts) added up to an exponent of three, I took the root of the left side and was left with an odd number.As I plugged in everything (with negative acceleration), I got a time of 0.58s. Am I rearranging the formula the wrong way then?
  5. Sep 13, 2007 #4


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    YEah, that isn't right... careful with the algebra... can you show all your steps... I can poit where it's going wrong.

  6. Sep 13, 2007 #5
    Ok, I thought I was doing something wrong but I wasn't sure what.

    d = v1t + (1/2)at^2
    2d/v1 = t + at^2
    2d/v1(a) = t + t^2

    Sorry if my algebra is a little rusty, I haven't touched it in a long time. I figure that it is either step 2 or 3 where I went wrong, but I'm not sure of any other way of doing it.
  7. Sep 13, 2007 #6


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    Yeah, the second step is wrong... multiply both sides by 2... do that as one step... then if you divide by v1... you have to divide everything by v1 (all 3 terms)...

    you don't really have to rearrage the equation... just plug in the numbers and solve for t... it's a quadratic equation.
  8. Sep 13, 2007 #7
    I'm still not sure what you mean by multiplying both sides by 2 and doing it as one step. Would the next step be:

    2d = 2(v1t + 1/2at^2)
    2d = 2v1t + at^2
    2(4) = 2(4)t + (-9.8)t^2

    I'm sorry, but I have forgotten quite a bit about algebra and can't remember how to specifically rearrange.
  9. Sep 13, 2007 #8


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    Yes, that's right. And if you want to divide by v1... you'd have to divide all 3 terms by v1...

    Do you remember the quadratic formula? Simplify the above equation...
  10. Sep 13, 2007 #9

    Ok, I just wanted to make sure I was doing it correctly first. I had forgotten that pulling a single term (from 1/2a) wouldn't really work out too well.

    2d = 2v1t + at^2
    8 = 8t + (-9.8)t^2
    (-9.8t^2) + 8t - 8 = 0

    However, if I plug that into the Quadratic Formula, I would get a negative under the root sign?
  11. Sep 13, 2007 #10


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    Ah yes... good catch... the d should be -4 (we're taking up positive, down negative)... because it's a displacement downwards of 4m...
  12. Sep 13, 2007 #11
    Hrmm, I thought the displacement wouldn't matter, but clearly I have to give everything a direction from a reference point...

    From here, I would plug it into the Quadratic Formula:
    (-9.8t^2) + 8t + 8 = 0
    b^2 - 4ac
    8^2 - (4)(-9.8)(8)
    ROOT-> 377.6

    The equation is then:

    X= [(-b) +- (19.43)]/2(-9.8)
    = [(-8) +- (19.43)]/(-19.6)
    = 1.4s

    Ahhh woooww.... Thanks so much for your help, I wouldn't have been able to do it if not for your guidance. I can't believe I forgot so much about basic algebra, I suppose it's time for a review.

    Once again, thanks a lot for your help. :)
  13. Sep 13, 2007 #12


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    No prob... with a little practice your algebra will back up to speed easily.
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