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Kinematics - Calculating Distance

  1. Oct 11, 2013 #1
    Question: "A man steps from the top of a tall building. He falls freely from rest to the ground, a distance of h. He falls a distance of h/4 in the last 0.800 s of his fall. Neglecting air resistance, calculate the height h of the building.

    I've attempted the solution in a number of different ways, each of which give different answers.

    Solution:
    I have labelled the initial time (i.e. at the instant before he begins to fall) as ##t_{0}##, the time at which he has traveled ##\frac{3}{4}## of the total distance h as ##t_{1}## and the time at which he has traveled the full height h as ##t_{2}##.

    Using the equation ##x = x_{0} + v_{0}t + \frac{1}{2}a_{x}t^{2}##, where ##x = h##, ##t = t_{2}## and ##a_{x} = g##.
    ##x_{0} = 0## as no distance has been traveled whilst the man is still at the top of the building
    ##v_{0}t = 0## as the initial velocity at the top of the building is zero

    Therefore -

    ## h = h_{0} + v_{0}t_{2} + \frac{1}{2}gt_{2}^{2}##
    ## h = 0 + 0 + \frac{1}{2}gt_{2}^{2}##
    ## h = \frac{1}{2}gt_{2}^{2}##

    Re-arranging to give -

    ##t_{2} = \sqrt{\frac{2h}{g}}##

    ##(t_{2} - t_{1})## = the time to travel ##\frac{1}{4}h = 0.800 s##
    ##(t_{1})## = the time to travel ##\frac{3}{4}h = \sqrt{\frac{2h}{g}} - 0.800 s##

    Substituting the value of ##t_{2}## for ##h## in the latter of these equations gives -

    ##\frac{3}{4}\sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}} - 0.800 s##
    Therefore 0.800 must be equivalent to ##\frac{1}{4}\sqrt{\frac{2h}{g}}##
    ##0.800 = \frac{1}{4}\sqrt{\frac{2h}{g}}##
    ##3.20 = \sqrt{\frac{2h}{g}}##
    ##h = \frac{3.20^{2}g}{2}##
    ##h = 50.2 m##

    I believe that my mistake is in assuming that ##(t_{1}) = \sqrt{\frac{2h}{g}} - 0.800 s## is equivalent to ##\frac{3}{4}h\sqrt{\frac{2h}{g}}##, as the man's velocity increases with constant acceleration and therefore ##\frac{3}{4}## of ##t_{2}## does not correspond to having traveled ##\frac{3}{4}h##.

    However, I'm uncertain how else to approach the problem. I have tried other methods involving heavier use of the equations of motion with constant acceleration and have twice calculated h as ≈ 12 m, but this would not be a tall building! (I realise that the actual answer does not always have to accurately reflect the real-world scenario described in the question, but I believe I should use it as guidance in evaluating my answer.)

    Any help and/or advice would be greatly appreciated.
     
    Last edited: Oct 11, 2013
  2. jcsd
  3. Oct 11, 2013 #2
    Yes, exactly. Up to the point of this assumption everything was correct. And you even know the reason. solve for ##t_{1}## just like you solved for ##t_{2}## but this time take the height to be ##\frac{3}{4}h##. Now, substitute this value and solve. You'll arrive at the answer.

    P.S. : ##t_{1}=\sqrt{\frac{3h}{2g}}##
     
  4. Oct 11, 2013 #3
    Hi, thankyou for your response.

    I had approached the problem this way previously, but for some reason or another I discarded it.
    I have tried to run it through and have calculated what I think is a reasonable answer -

    ##t_{2} = \sqrt{\frac{2h}{g}}##
    ##t_{1} = \sqrt{\frac{3h}{2g}}## (derived from the equation of motion used previously using ##\frac{3}{4}h## as ##x##)
    ##(t_{2} - t_{1}) = 0.800 s##
    ##t_{2} - (t_{2} - t_{1}) = t_{1}##
    ##\sqrt{\frac{2h}{g}} - 0.800 s = \sqrt{\frac{3h}{2g}}##
    ##{\left(\sqrt{\frac{2h}{g}} - 0.800 s\right)}^2 = {\left(\sqrt{\frac{3h}{2g}}\right)}^2##
    ## \frac{2h}{g} - 1.6\sqrt{\frac{2h}{g}} + 0.64 = \frac{3h}{2g}##
    ##\frac{h}{2g} - 1.6\sqrt{\frac{2h}{g}} + 0.64 = 0##

    Solving the quadratic equation using the following substitutions -

    ##u = \sqrt{2h}## and ##u^{2} = 2h##
    ##\frac{1}{4g}u^{2} - \frac{1.6}{\sqrt{g}}u + 0.64 = 0##

    Using the quadratic formula -
    ##x = -\frac{b\pm\sqrt{b^{2}-4ac}}{2a}##

    ##u \approx 18.7026##
    ##u \approx 1.3429##
    ##u = \pm\sqrt{2h}##
    ##h = 175 m##

    I hope I haven't made any errors, please correct me where appropriate. I realise that I should try to keep the value of u as an exact number for use in further calculation, but I kept making mistakes when trying to input values in my calculator and found it easiest to right down the values of a, b and c to high precision for use in the formula, as the final answer is required to only three significant figures.

    Thanks again for your input :smile:
     
  5. Oct 12, 2013 #4
    Yes, its correct. And you have managed to separate out the real solution from the unreal one.

    P.S. : Separating the roots is done by checking the values in the original equation involving ##t_{1}## and ##t_{2}##
     
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