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Acceleration due to Gravity

  1. Feb 24, 2008 #1
    1. A flowerpot is dropped from the balcony of an apartment, 28.5 m above the ground. At a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0 m above the ground. The initial velocity of the ball is 12.0 m/s [down]. Does the ball pass the flowerpot before striking the ground? If so, how far above the ground are the two objects when the ball passes the flowepot?

    2. a = v/t
    all other variations of this constant acceleration formula.
    d = v(initial) x t + 1/2 x a x t^2

    3. 1. I used the above formula (The one with the 'd') to find the time of flight for the ball.
    2. I used the same formula to find the distance flowerpot fell in 1 second.
    3. I subtracted it from the overal distance to find the position of the flowerpot when the ball was dropped.
    4. I used that position to find the velocity at that time.
    5. Using that 'd' formula again I find the time it takes for the flower pot to fall after the ball is dropped. (Not so sure about this line anymore)
    6. No idea how to find the displacement. All I know is that the displacement must equal at that point...
  2. jcsd
  3. Feb 24, 2008 #2


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    Staff: Mentor

    One could solve for the time, t, that the flower pot takes to reach the ground. The ball must then reach the ground in less than t-1 s in order to pass the flower pot. In this case, t is the differential time or period of time.

    Here are some useful notes on trajectories and freefall.
  4. Feb 24, 2008 #3
    As I have said, I've already solved for the times. What I don't know how to do is the seconds part. It's asking me for the distance from the ground when the ball passes the flowerpot.

    Hurrah for large quantaties of formulea, but I have no idea which ones I should use, or what to even do with them. Every time I make something equal I still have two variables left over and I don't see any connection between any of the values except time. And when I tried that one it failed miserably.

    So hurrah and thankyou for that wonderful website, but I have a Physics textbook, a different physics book, and the internet. What I really need is some help and advice, not genereal knowledge.
  5. Feb 24, 2008 #4


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    Staff: Mentor

    Which one hits the ground first? If the flower pot strikes the ground first, then the ball does not pass, and therefore there is no height at which the ball passes the flower pot.

    Note that the ball starts 1 s after the flower pot.
  6. Feb 24, 2008 #5
    The ball hits first. It takes the ball 1.38 seconds to hit the ground. While it takes the flowerpot 2.75 seconds overall.
  7. Feb 24, 2008 #6


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    Gold Member

    OK that's the easy part. Now explain why, on its shortlived journey, the flowerpot was heard to say "Oh no. Not again."
  8. Feb 24, 2008 #7
    Hitchhikers reference eh? The other object should be the falling whale.
  9. Feb 24, 2008 #8
    I hate you both. Though I did draw a picture with it saying that as it falls when I began the question.
  10. Feb 25, 2008 #9


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    Staff: Mentor

    Well one wants to find h(t), the position above the ground where the ball is the same height at the flower pot.

    Let hf(t) = 28.5 m - df(t), where df(t) is the displacement of the flower pot, and

    hb(t) = 26 m - db(t), where db(t) is the displacement of the ball, then

    let hf(t) = hb(t), solve for t, then use either equation to solve for h(t).

    Please, write the equations for df(t) and db(t).

    Note that at 1 s, db(t = 1s) = 0.

    Here is a good reference on the derivation of the equation of linear motion at constant acceleration.

    The problem at hand deals with one event and another delayed event.
    Last edited: Feb 25, 2008
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