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Spiderman Free fall problem

  1. Jan 31, 2006 #1
    EDIT: Thanks Hootenanny and Doc Al for looking and trying the problem. I figured out my mistake [tex]h_2 = h - \frac{h}{n}[/tex] not [tex]h_2 = 1- \frac{h}{n}[/tex]

    I've been working on this for a sick amount of time, please help me figure out what I'm doing wrong...

    The question is:

    Spiderman 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 [tex] \frac{h}{n} [/tex] in the last interval of time of ∆t of his fall. What is the height h of the building?

    There's probably a much easier way to approach this but this is what I did (yea, there's room for alot of error):

    I made [tex]h_1 = \frac{h}{n}[/tex] and [tex]h_2 = 1- \frac{h}{n}[/tex]

    [tex]V_{final}^2 = 2gh_2[/tex]
    [tex]V_{final} = \sqrt{2gh_2}[/tex]

    since Vfinal from 0 to 1- h/n is equal to Vinitial from h/n to h...

    0 = [tex]h_1 + \sqrt{2gh_2} \Delta t - 0.5g\Delta t^2\\[/tex]
    [tex]0.5g\Delta t^2 - h_1 = \sqrt{2gh_2} \Delta t[/tex]
    [tex]0.5g\Delta t - \frac{h_1}{\Delta t} = \sqrt{2gh_2}[/tex]
    [tex](0.5g\Delta t - \frac{h_1}{\Delta t})^2 = 2gh_2[/tex]
    [tex]0.25g^2\Delta t^2 - gh_1 + \frac{h_1^2}{\Delta t^2} = 2gh_2[/tex]

    I set [tex]\beta = 0.25g^2\Delta t^2[/tex] and plug in the values for h1 and h2

    [tex]\beta - \frac{g}{n} h + \frac{1}{\Delta t^2 n^2} h^2 = 2g - \frac{2gh}{n}[/tex]
    [tex](\beta - 2g) + \frac{g}{n}h + \frac{1}{\Delta t^2 n^2} h^2 = 0[/tex]

    [tex]Ah^2 +Bh +C = 0[/tex]
    [tex]A = \frac{1}{\Delta t^2 n^2}[/tex]
    [tex]B = \frac{g}{n}[/tex]
    [tex]C = 0.25g^2\Delta t^2 -2g[/tex]

    and then I solved the quadratic. um, extremely incorrect...I was able to check my answer because the same problem was written as:

    "Spiderman 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 interval of time of 1.0 sec of his fall. What is the height h of the building?"

    in another source and the answer is h = 270m

    Thanks in advance :)
     
    Last edited: Jan 31, 2006
  2. jcsd
  3. Jan 31, 2006 #2

    Hootenanny

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    Staff Emeritus
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    Gold Member

    Try using [tex] v^2 = u^2 + 2as [/tex]
    where [itex] v = final velocity, u = initial velocity, a = acceleration = g, s = displacement = h [/itex]
    using your working above you could sub [tex] v^2 = 2gh [/tex] into the above equation.
    Edit: Just a note to say that this solution neglects air resistance.
     
    Last edited: Jan 31, 2006
  4. Jan 31, 2006 #3
    Thanks for replying.

    That is the equation I used in the first step; should I be using it again somewhere else?

    I guess I forgot to mention that we have only learned one dimensional motion and vectors so air resistance is still being ignored in all the problems.
     
  5. Jan 31, 2006 #4

    Doc Al

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    Staff: Mentor

    Try using the kinematic formula relating distance and time:
    [tex]s = 1/2 a t^2[/tex]
     
  6. Jan 31, 2006 #5
    Thanks for the suggestion. I actually just figured out what I did wrong ( [tex]h_2 = h - \frac{h}{n}[/tex] not [tex]h_2 = 1 - \frac{h}{n}[/tex]). :P
     
  7. Jan 23, 2008 #6
    My question:
    Spiderman 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/ 3 in the last interval of time of 1.0 s of his fall.

    I technically have the same problem, however a different n value. I am trying the solution here (obviously with 3 not 4) and can not find the correct answer... any suggestions? thanks alot
     
  8. Jan 23, 2008 #7
    Scratch that, I was able to rearrange it and understand.

    For a h/3 with a time of 1 sec the answer is 150...
     
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