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Rubber Duckie Drop

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A hot air balloon containing an AP Physics student ascends vertically at a constant speed of 8 m/s. While ascending, he accidentally drops his rubber duckie from the gondola of the ballong. Seven seconds after it is dropped the duckie bounces off a roof at point A and hits the ground at point B as shown in the sketch at the right.

    A. Find the distance from the point where the duckie is dropped to point A
    B. Find the speed of the sweet little rubber duckie just before it hits the roof

    Suppose that immediately after bouncing off the roof, the velocity of the rubber duckie is 12m/s 37 degrees BELOW the horizontal and that point B is horizontal distance of 24 meters from point A.

    C. Find how far point A is above point B
    D. Find the duckie's time of flight between points A and B
    E. Find the Velocity of the duckie the instant before it strikes the ground at Point B


    2. Relevant equations
    V = Vo+at
    V^2 = Vo^2 + 2ad
    d = VoT + 1/2at^2

    3. The attempt at a solution
    Currently just filled out my DxVxT and VoyVyadyt to the following:
    Voy: 8m/s
    Vy: ?
    a: -9.81m/s^2
    dy:?
    T:?
     
  2. jcsd
  3. Sep 15, 2015 #2

    berkeman

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    Staff: Mentor

    You haven't really shown much of an attempt at the solution yet. What is the equation for the vertical position of the ducky as a function of time?
     
  4. Sep 15, 2015 #3
    Would that not be the VoyVyaDyT?
    Or maybe it is DxVxT. I'm really unsure right now; need some info and I can solve the rest.
     
  5. Sep 15, 2015 #4

    berkeman

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    Staff: Mentor

    Sorry, I have no idea what any of that means.

    You have the beginning of the right equation when you wrote "d = VoT + 1/2at^2", but it is incomplete. What is the full equation for y(t) in terms of y_0, Vy_0, acceleration and time...?
     
  6. Sep 15, 2015 #5
    Ah so would it be "d = (8.0)t + 1/2(-9.81)(t)^2"
     
  7. Sep 15, 2015 #6

    berkeman

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    Staff: Mentor

    Closer, but still some issues.

    First, I would use y(t) on the LHS of the equation instead of d. y(t) is much more descriptive of what is going on (the ducky is moving vertically with time).

    Second, you have left off the initial y position of the ducky, y(0). Where does that go in the equation?

    Once you get those fixed, you have an equation for the vertical position y(t) of the ducky with respect to time. What else are you told in the problem statement? How can you use those facts and this equation to come up with the first answer to the problem? :smile:
     
  8. Sep 15, 2015 #7
    Hmm..
    So the initial position would be 0 so the equation would come out to be:
    0 = (8.0)t + 1/2(-9.81)t^2
    Correct?
     
  9. Sep 15, 2015 #8

    berkeman

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    Staff: Mentor

    No. Can you explain why you wrote that equation in response to my hints?
     
  10. Sep 15, 2015 #9
    So what I initially planned to do first was to find the distance of the balloon itself, leaving the bigger picture apart. I thought that if I found how high the balloon is, I could find the duck falling and speed easily.
    This was my plan:
    Balloon: Vo = 8.0 m/s
    Acceleration = -9.81m/s^2
    However, I need one more factor to begin solving for any variable. I thought that since it took the duck 7 seconds to fall, I could just make the time of the balloon as 7 seconds.
    Where did I go wrong?
     
  11. Sep 15, 2015 #10
    Sorry, I'm not sure what some of the terms you state actually mean.
     
  12. Sep 15, 2015 #11

    berkeman

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    Staff: Mentor

    This is the equation you need to write and fill out. It is commonly used in problems like this one:

    [tex]y(t) = y_0 + {Vy_0}t + \frac{1}{2}{a_y}t^2[/tex]
     
  13. Sep 15, 2015 #12
    Nevermind, I have solved all of the problems. The equations were quite easy to be honest, just couldn't make out a mental picture.
    Thanks for the help!
     
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