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Homework Help: Kinematics motion

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A balloon is rising at 10m/s when its passenger throws a ball straight up at 12m/s relative to the balloon . How much later does the passenger catches the ball?

    2. Relevant equations

    3. The attempt at a solution

    vf^2 - vi^2 = 2g(yf-yi)

    0ms^-1 - (22ms^-1)^2 = 2(-9.8ms^-2)(yball)
    yball = 24.7m

    0ms^-1 - (10ms^-1)^2 = 2(-9.8ms^-2)(yballoon)
    yballoon = 24.7m
    Last edited: Dec 13, 2013
  2. jcsd
  3. Dec 13, 2013 #2
    Assume that the velocity of the balloon is constant.
  4. Dec 13, 2013 #3

    You're right since the ball is thrown up relative to the balloon.
    Last edited: Dec 13, 2013
  5. Dec 13, 2013 #4
    The balloon is not a projectile, there are other forces acting on it. That does not mean, in general, that the acceleration is zero, but since you are not given any, the only sane assumption is that the acceleration is zero.
  6. Dec 13, 2013 #5
    yf - yi = vit + 0.5gt^2


    yball = (12ms^-1 + 10ms^-1)t + 0.5(-9.8ms^-2)t^2


    yballoon = 10ms^-1 t + 0.5(oms^-2)t^2
    y balloon = 10ms^-1

    yballoon = yball

    10ms^-1 t + 0.5(oms^-2)t^2 = 10ms^-1
    t = 2.4s
  7. Dec 13, 2013 #6
    There is a simpler approach. Since the balloon is not accelerated, everything can be computed with regard to the balloon as if it was stationary. So that basically means the projectile goes up at 12 m/s. At some time t it will reach a maximum elevation (from the balloon). This time can be easily computed from the condition that the (relative) velocity becomes zero. Going down takes the same time.
  8. Dec 13, 2013 #7
    So is 2.4s the right answer?
    It appears more logical if I multiply 2.4s by 2.
    Last edited: Dec 13, 2013
  9. Dec 13, 2013 #8


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    You correctly calculated 2.4s as the time when ball and balloon would meet again. If you had calculated the time to when the relative velocity were zero, then you would have got 1.2s and would need to double for the second half.
    In fact, a slightly simpler calculation than the one you did is to say that they will meet again when the relative velocity is equal and opposite to its initial value, giving gt = 24m/s.
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