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Dollar bill Free fall problem

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Your friend makes a wager with you. He holds a $50 bill between your thumb and finger, and says you can keep the $100 bill if you can catch it when it drops. The bill is 16cm long. Using your reaction time do you catch the bill?



    2. Relevant equations



    3. The attempt at a solution
    Rxn time: 0.20 s
    I assume acceleration and velocity are relevant, but I don't know how to solve it. We solved a similar example in class, but I can't remember.
     
  2. jcsd
  3. Feb 6, 2014 #2
    The approximate distance fallen and the acceleration due to gravity are both known. Hence one can calculate an approximate value for the reaction time and then a decision can be made if this is possible.
     
  4. Feb 6, 2014 #3
    d=16cm
    a=9.8 m/s^2 (would this be negative?)
    t=?
     
  5. Feb 6, 2014 #4
    Would I use this formula: y(t) = vi*t - 1/2gt^2

    0.16m = 0m/s*t - 1/2*9.8 m/s^2*t^2
    0.16 m=4.9 m/s^2 *t^2
    0.16 m/4.9 m/s^2 = t^2
    then square it?
     
  6. Feb 6, 2014 #5
    Note that units of distance have to be consistent.
     
  7. Feb 6, 2014 #6
    Yes that is the correct formula except that I would have put (1/2) instead of 1/2.
     
  8. Feb 6, 2014 #7
    If you square your answer you get a value for t[itex]^{4}[/itex] and not for the value of t.
     
  9. Feb 6, 2014 #8
    Oops, I typed the wrong formula. In my text, it shows that it is y(t)=yi +vi*t -(1/2)gt^2

    What do i do about the yi?
     
  10. Feb 6, 2014 #9
    It is OK. Just take y[itex]_{i}[/itex] to be 0.
     
  11. Feb 6, 2014 #10
    That is the distance fallen is going to be measured from the top of the dollar bill.
     
  12. Feb 6, 2014 #11
    y(t) = vi*t - 1/2gt^2

    0.16m = 0m/s*t - 1/2*9.8 m/s^2*t^2
    0.16 m=4.9 m/s^2 *t^2
    0.16 m/4.9 m/s^2 = t^2
    t=0.18 s

    so then theoretically i could not catch the dollar bill? right?
     
  13. Feb 6, 2014 #12
    The calculation is ok.

    The final decision is for you to decide!
     
  14. Feb 6, 2014 #13
    Kinematics

    To clarify, I think there are some additional conditions: your thumb and index finger are aligned with the bottom of the bill, and you cannot move your hand.
    When your friend drops the bill, you must have pinched your thumb and index finger together at or before the time the top of the bill falls to their height. In other words, your reaction time must be less than or equal to the time it takes the bill to fall 16 cm.
    We know that objects at or near Earth's surface experience an acceleration of ~9.8 meters per second every second toward the Earth's center of mass. Since the bill remains close to Earth's surface throughout its fall, we may assume acceleration is a constant -9.8 m/s2 as long as we define the upward direction to be positive. Therefore, we may define a function that relates the acceleration of this bill to the elapsed time, a(t) = -9.8. From this, we can antidifferentiate a(t) to find a family of functions that relate the velocity of the bill to time, v(t) = -9.8t + vi, where vi is the initial velocity of the bill. Since it is at rest before being dropped, vi = 0, and v(t) = -9.8t. Antidifferentiate v(t) to find a family of functions that relate the position of the top of the bill to time, p(t) = -4.9t2 + hi, where hi is the initial height of the top of the bill. If you let your hand be the reference point, then the hi = .16 m, and p(t) = -4.9t2 + .16. We want to know the elapsed time when the top of the bill has reached your hand, so we set p(t) = 0.
    -4.9t2 + .16 = 0
    t = [itex]\sqrt{}(.16/4.9)[/itex] ≈ .18 seconds, so with a reaction time of .2 seconds, the bill will fall through your fingers before you can react, and your friend will win the bet.
     
  15. Feb 6, 2014 #14
    One final difficulty to be solved.

    Reading the first post one wonders how the original bill of $50 changes into one of $100 when it is falling!
     
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