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Homework Help: Falling Objects (With air resistance)

  1. Jan 29, 2010 #1
    So, I've been having a bit of an argument with a guy about falling, identical items, with varying masses.

    1. The problem statement, all variables and given/known data

    For the sake of the argument, we've been using the example of two identical jugs, one filled with water, and the other without. They are dropped at exactly the same time. I believe the one filled with water will hit the ground first, due to air resistance. He believes they will hit at the same time. I agree that they will, in a vacuum.

    2. Relevant equations

    Force of Air Resistance = Constant
    Fg = m * g

    3. The attempt at a solution


    In situations in which there is air resistance, more massive objects fall faster than less massive objects. But why? To answer the why question, it is necessary to consider the free-body diagrams for objects of different mass. Consider the falling motion of two skydivers: one with a mass of 100 kg (skydiver plus parachute) and the other with a mass of 150 kg (skydiver plus parachute). The free-body diagrams are shown below for the instant in time in which they have reached terminal velocity.

    I'm not entirely sure why we're still having this argument, but he's very insistent, so I'd like a final answer from someone who knows this firsthand.
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 29, 2010 #2


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    Yes, the heavier jug will hit the ground faster. There's absolutely no room for an argument to the contrary. To take an extreme case, a paper jug would take seconds to fall to the ground from eye level.
  4. Jan 29, 2010 #3
    As the (possibly mentally defective) friend from morcams above post, I'd like to ensure I'm totally wrong by asking for a second opinion.
  5. Jan 29, 2010 #4


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    Until the second opinion comes, I'll expand a bit on my answer. The force of gravity is mg, and air resistance is kAv^2. k is a constant, A is the cross-sectional area, and v is speed. Newton's second law says F=ma, so mg-kAv^2 = ma and a=g-kAv^2/m. Solving this equation will yield a position vs. time equation, but it's not necessary to go that far. Just examining the equation will do.

    Go through this in your head: A jug is dropped. Gravity accelerates it at 9.8 m/s: that's the "g" in the equation a=g-kAv^2/m. If the mass is tiny, kAv^2/m would be huge, and acceleration would immediately drop. If the mass is high, kAv^2/m would be tiny, and acceleration would decrease slowly.

    So the conclusions are:

    (1) At every point in time during the fall, the heavier jug has a greater acceleration than the light one.
    (2) At every point in time during the fall, the heavier jug has a greater speed than the light one.
    (3) At every point in time during the fall, the heavier jug is closer to the ground than the light one.
  6. Jan 29, 2010 #5
    I think perhaps I should have drunk a little less at the weekends while at collage, somehow I managed to have an hour argument over something so very obvious.
    I was so sure I was right, I suppose this is how wars get started!
    I'm going to accept my utter stupidity and ignorance in this matter and thank you kindly for your reply.

    Now all I have to do is survive morcams jokes :/
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