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Free-Fall problem

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    hot air balloon travelling vertically upward at 2.2 m/s. at 25 m above ground, releases a package. Solve for t.


    2. Relevant equations
    d=v(initial)t + 0.5(a)t2


    3. The attempt at a solution

    25m = [0.0 m/s(t)] + ½(9.8m/s2)t2
    25/4.9 = t2
    t=2.26 seconds

    My question is, does the upward movement of the balloon impact the time that the package is in the air? If so, would it be:

    25 = -2.2t +0.5(9.8)t2
    t= 2.49 seconds
     
  2. jcsd
  3. Sep 9, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Absolutely. The fact that the package was riding with the balloon tells you its initial velocity.
     
  4. Sep 9, 2007 #3
    Yes, (your second attempt, although I didn't check the arithmetic)
    The ball would initially be moving at the same velocity as the air balloon all though gravity would immediately cause it to slow down to a stop, then to speed up on its way down. For some examples to help you get this concept down, consider a jet dropping a bomb - the bomb initially has the same forward velocity as the jet - it doesn't simply fall straight down from the point of release. Instead, it follows a parabolic path (ignoring air resistance and the curvature of the earth.) Another example would be to drop a dense object out the window of a moving car (at not too high of a speed such that air resistance becomes a huge factor.) The object will remain beside the drivers door while it falls to the ground.
    For another example, see this video of a cart shooting a ball vertically (with respect to the cart) - notice that the horizontal velocity of the ball is the same as the horizontal velocity of the cart at the moment it leaves the cart.
    http://www.physics.umd.edu/lecdem/services/demos/demosc2/c2-25.htm

    edit: (Wow, I'm slow! DocAl is too quick for me.)

    What would trick people into thinking that the ball being dropped from the balloon is immediately going in the downward direction is simply because they're using the balloon as a frame of reference, rather than the ground. Plus, the low velocity of the balloon means the ball will reach a velocity of 0 m/s in only about a fifth of a second. If it were a piece separating from a rocket at a high velocity, it might be more noticeable that with respect to the ground, the piece continues moving vertically upward until it slows to a stop.
     
    Last edited: Sep 9, 2007
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