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1D kinematics - finding g from time traveled

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data
    This is from Kleppner, problem 1.12 in the 2010 edition, if you want to see the diagram.

    The acceleration of gravity can be measured by projecting a body upward and measuring the time that it takes to pass two given points in both directions. Show that if the time the body takes to pass a horizontal line ##A## in both directions is ##T_A##, and the time to go by a second line ##B## in both directions is ##T_B##, then, assuming that the acceleration is constant, its magnitude is
    [tex]g = \frac{8 h}{T_A^2 - T_B^2},[/tex]where ##h## is the height of line ##B## above line ##A##.

    2. Relevant equations
    ##y = y_0 + \dot{y}_0 t - \frac12 g t^2##

    ##\dot{y} = \dot{y}_0 - g t##

    3. The attempt at a solution
    Since we're only interested in what happens above height ##A##, I'm going to make that my zero height, so the total flight time is ##T_A##, making the time for the body to reach its max height ##T_A / 2##. From that I can get the initial velocity from ##\dot{y} = 0 = \dot{y}_0 - g ( T_A / 2 )##, giving me
    [tex]\dot{y}_0 = \frac{g T_A}{2}.[/tex] So now I've got an equation for height
    [tex]y = \frac{g}{2} ( T_A t - t^2 ),[/tex]which I can use to find an expression for the max height by plugging in ##t = T_A / 2##:
    [tex]y_{\text{max}} = \frac{g}{2} \left( T_A \frac{T_A}{2} - \frac{T_A^2}{4} \right) = \frac{g T_A^2}{8}.[/tex]
    I also want to know how long it takes to get to ##B##; I'll call that ##t_h##. Plugging that into my height equation,
    [tex]h = \frac{g}{2} ( T_A t_h - t_h^2 ),[/tex] and solving for ##t_h## I get
    [tex]t_h = \frac{T_A}{2} \pm \frac12 \sqrt{T_A^2 - \frac{8 h}{g}},[/tex] the minus term being the time it takes to reach ##B## the first time and plus term the time it takes to reach ##B## the second time.

    Now I'm going to focus on ##B##. I can get another expression for the max height by using time ##T_B/2## and initial velocity ##\dot{y}_B = \dot{y}_0 - g t_h##. Plugging in everything I have so far,
    [tex]\dot{y}_B = \frac{g T_A}{2} - \left( \frac{g T_A}{2} - \frac12 \sqrt{T_A^2 - \frac{8 h}{g}} \right) = \frac12 \sqrt{T_A^2 - \frac{8 h}{g}}.[/tex]And the equation for max height is
    [tex]y_{\text{max}} = \frac{T_B}{2} \frac12 \sqrt{T_A^2 - \frac{8 h}{g}} - \frac{g}{2} \frac{T_B^2}{4} = \frac{T_B}{4} \sqrt{T_A^2 - \frac{8 h}{g}} - \frac{g T_B^2}{8} = \frac{g T_A^2}{8}.[/tex]Simplifying,
    [tex]T_B \sqrt{T_A^2 - \frac{8 h}{g}} = \frac{g}{2} ( T_A^2 + T_B^2 ).[/tex]
    Here's where I'm stymied. If there's a way to solve that for ##g##, I don't know it, and the magic equation-solving machines don't spit out anything nice, certainly not the given solution. So I suppose I've gone astray somewhere. Help?
     
  2. jcsd
  3. Apr 13, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks overly complicated.

    Hint: View the body as it falls, starting from the highest point.
     
  4. Apr 13, 2013 #3
    Solved it. Thanks.
     
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