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Homework Help: Basic kinematics of a balloon rising

  1. Feb 3, 2008 #1
    Q- A balloon is rising at 10m/s when its passenger throws a ball straight up at 12m/s .How much later does the passenger catch the ball?

    How would I approach this problem ? I am terrible at physics when it comes to doing the first basic step on a problem. If anyone can point me in the right direction it would be great.

    Thank you!

    Edit: units.
    Last edited: Feb 3, 2008
  2. jcsd
  3. Feb 3, 2008 #2
    Think about doing the problem with respect to the speed of the balloon. (i.e. the ball is thrown at [tex]2 \tfrac{m}{s}[/tex] with respect to the balloon.)

    And, in future, include units. It's really very inconvenient to have to figure out whether you meant the balloon started rising at t=10s or whether its speed at t=0 is [tex]10 \tfrac{m}{s}[/tex].
  4. Feb 3, 2008 #3
    what equaton should I use--should I take gravity into consideration?
  5. Feb 3, 2008 #4
    Ok, I have delta x = 0 = v_initial*t - 1/2*9.8*t^2.

    which gives me t = 0 and t= .5s .
    I used 2m/s for the velocity.

    Is this correct.
  6. Feb 3, 2008 #5
    You should have gotten [tex] t = .41s.[/tex] Maybe you rounded wrong?
  7. Feb 3, 2008 #6
    Also, you could have solved the problem as a system of equations:
    y &= 10t\\
    y &= 12t - \frac{1}{2}gt^2

    Both ways result in the same answer.
    Last edited: Feb 3, 2008
  8. Feb 3, 2008 #7
    the online homework site is saying that .41s (two sig figs) is not the right answer. Are u sure it is correct?
  9. Feb 3, 2008 #8
    I can't see how it wouldn't be.
  10. Feb 3, 2008 #9
    Do you use webassign?
  11. Feb 3, 2008 #10
    no, i use mastering physics. I copied and pasted the question, but it is giving me try again. what should I do?

    fox, why didn't u take gravity into account for y=10t?
  12. Feb 3, 2008 #11
    well, I was assuming that it meant that the balloon was rising at constant speed. But, hey, why don't you try it with gravity? Maybe that'll be right. At least I hope so.
  13. Feb 3, 2008 #12
    but then the gravity will cancel out with the other gravity term.
  14. Feb 3, 2008 #13
    Hmm. you're right. Rrrr. Why don't you email your teacher. See if maybe mastering physics is wrong.
  15. Feb 4, 2008 #14
    What do you think of this:

    speed of the ballon along vertical direction v = 10 m / s
    speed of the ball along vertical direction u = 12 m / s
    relative speed of the ball with respect to nallon V = 12 - 10 m / s
    = 2 m / s
    time taken to catch the ball by the passinger t = 2V /g
    = 0.804 s
  16. Feb 4, 2008 #15
    2V/g = 2*2/9.81 = 0.41 s
  17. Feb 4, 2008 #16
    That's what I got, but according to the website thingy he's using, that's not correct.
  18. Feb 4, 2008 #17

    gravity should only 'apply' to the ball, since the gravity on the balloon is canceled by its upwards force..
    the balloon travels at a constant 10 m/s, the ball starts at 12 m/s with acceleration -9.8 m/s².
    Last edited: Feb 4, 2008
  19. Feb 4, 2008 #18
    OK, so I will talk to the ta and see what transpired...
  20. Feb 4, 2008 #19
    What about if I look at it relative to the ground:

    The answer would be
    d=10t for passenger,
    d=22t-4.9t^2 (22m/s relative to the ground)

    2.44s. Perhaps this is the answer--we will see. As much as I hate the long chapters and impossible problems on Halliday, at least it is easy to get a hold of the answers for that book.
  21. Feb 4, 2008 #20
    Hmm. I was thinking the values you were given were already with respect to the ground, but maybe only the balloon's speed was. In that case 2.44s would be the answer. But I'd talk to the ta anyway. The question should be clearer if that's so.
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