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Baseball dropped from building with air resistance

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A baseball (radius = 0.0366m, mass = 0.145kg) is dropped from rest at the top of the Empire State Building (height = 1250ft = 381m). Calculate (a) the initial potential energy of the baseball, (b) its final kinetic energy, and (c) the total energy dissipated by the falling baseball by computing the line integral of the force of air resistance along the baseball's total distance of fall. Compare this last result to the difference between the baseball's initial potential energy and its final kinetic energy.


    2. Relevant equations
    F = ma = -mg -cv
    P.E.=mgh
    [itex]c=1.1346\cdot 10^{-5}[/itex]

    3. The attempt at a solution
    For part a, it was easy to find the potential energy.

    mgh = (0.145kg)(9.8m/s2)(381m) = 541 J

    Part b is where I am stuck. I took down to be negative. For the force, we have

    [itex]F = ma = -mg - cv[/itex]

    since I was told by the professor to use the linear term rather than quadratic. Then we have

    [itex]ma=mv\frac{dv}{dx}=-mg-cv\Rightarrow \frac{mvdv}{-mg-cv}=dx[/itex]
    [itex]\Rightarrow \int \frac{mvdv}{-mg-cv}=\int dx[/itex]

    Where the limits of integration on the left are from [itex]v_0=0[/itex] to [itex]-v_f[/itex] and the limits of integration on the right are from [itex]x_0=381m[/itex] to [itex]x_f=0[/itex]. We now have

    [itex]\frac{-mv}{c}+\frac{m^2g\ln(mg+cv)}{c^2}[/itex] evaluated using the bounds mentioned above, this is the left hand side. The right hand side is [itex]x_f-x_0=-381m[/itex].

    Now we have
    [itex]\frac{mv_f}{c}+\frac{m^2g\ln(mg-cv_f)}{c^2}-\frac{m^2g\ln(mg)}{c^2}=-381[/itex]

    I tried solving this numerically just using Maple, but the value I got for the final velocity does not make sense when I compare my answer for the final kinetic energy to the back of the book. The back of the book has the final kinetic energy as 87 J. Using the equation for kinetic energy

    [itex]T=\frac{1}{2}mv^2[/itex]

    this means that the final velocity is 34, based on the final kinetic energy given in the back of the book. Which is far different from the final velocity of 86.39 that I get. Where am I going wrong?

    Note: I have not tried to do part c yet, this is just focused on part b.
     
  2. jcsd
  3. Sep 24, 2013 #2

    SteamKing

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    Is the drag force on the baseball acting in the same direction as the force due to gravity? According to your F = ma equation, the ball should fall faster with drag than without. Does that seem logical?
     
  4. Sep 24, 2013 #3
    The drag acts in opposition to the velocity (always in opposition to the direction of movement), so it should be acting in the opposite direction to the force due to gravity. However, in lecture, as something is falling the professor put [itex]mg[/itex] and [itex]cv[/itex] to be negative. So I'm wasn't sure what to do.
     
  5. Sep 24, 2013 #4
    Consider one of the directions to be positive. Let the downward direction be positive. mg acts downward and the resistive force upwards. Can you now apply F=ma?
     
  6. Sep 24, 2013 #5
    Ah ok, thank you. For F = ma = mg - cv. However, I just got back from class and apparently he was mistaken, we were supposed to use the quadratic term for the air resistance rather than the linear. I've got it now, thanks again.
     
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