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Introductory Physics Homework Help
Kinematics - Calculating Distance
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[QUOTE="Insolite, post: 4534362, member: 490825"] [B]Question:[/B] "A man steps from the top of a tall building. He falls freely from rest to the ground, a distance of [I]h[/I]. He falls a distance of [I]h/4[/I] in the last 0.800 s of his fall. Neglecting air resistance, calculate the height [I]h[/I] of the building. I've attempted the solution in a number of different ways, each of which give different answers. [B]Solution:[/B] 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 [I]h[/I] as ##t_{1}## and the time at which he has traveled the full height [I]h[/I] 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 [I]h[/I] as ≈ 12 m, but this would not be a tall building! (I realize 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. [/QUOTE]
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Kinematics - Calculating Distance
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