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Homework Help: Kinematics balloon problem

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data

    A balloon is rising vertically at a constant speed of 21 m/s. A stone is dropped from the balloon. The stone strikes the ground 10 seconds later. Assuming that air resistance is negligible and that acceleration on earth due to gravity is 9.8 m/s2, from what height is the stone dropped?

    2. Relevant equations

    v = u + at
    dv/dt = a
    dx/dt = v

    3. The attempt at a solution

    This is screwing with my head. It's an obvious displacement problem, but where would you set the origin, to which the displacement is relative?

    Here's my attempt:

    The acceleration of the stone at any time is 9.8, and because this is constant
    acceleration, v = u + at where a = 9.8 and u = 0 (because if you haven't dropped the stone yet, it must have a velocity of 0) .:. v = 9.8t

    Integrating that you get x = 9.8(t^2/2) + c .:. x = 4.9t^2 + c

    Here's where I run into trouble. WHERE IS THE ORIGIN?? Is it at the ground? At the balloon? At the height from which the stone is dropped? How do I define the displacement?

    I tried x = 0 at the ground .:. x = 0 when t = 10 .:. 0 = 4.9(100) + c

    .:. c = 490

    so x = 4.9t^2 + 490

    So the displacement at any time is equal to 4.9t^2 + 490.

    We know that the stone is dropped after 0 seconds, so the height from which the stone is dropped would be a height of 490 metres, according to what I came up with.

    If I'm wrong, could you please point out where I went wrong? In my approach I completely ignored the fact that the balloon rises at 21m/s. Is that part relevant?
  2. jcsd
  3. Dec 20, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Taking the orgin to be at a height of zero is perfectly OK. Ignoring the initial speed of the stone is not OK--it's not zero! (What's the initial speed of the stone?)

    Choose a coordinate system and sign convention, such as "up is positive; down is negative". Note that the acceleration would be -9.8 m/s^2.
  4. Dec 20, 2006 #3
    Thanks a lot. I'll revise what I did, then:

    The initial speed of the stone is 21m/s, if I'm re-evaluating correctly. The acceleration is -9.8m/s^2.

    The velocity is -9.8t + c ==> at t = 0, v = 21 ==> 21 = c so:

    v = -9.8t + 21

    Integrate to get X (displacement) = -9.8(t^2/2) + 21t + c
    = -4.9t^2 + 21t + c

    x = 0 when t = 10 ==> 0 = -490 + 210 + c ==> c = 280

    x = -4.9t^2 + 21t + 280

    At x = h, t = 0: h = 280

    280 metres?

    If it's correct, thanks a lot. If it's not, thanks anyway.
  5. Dec 20, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good.
  6. Dec 20, 2006 #5
    Okay, awesome.
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