(adsbygoogle = window.adsbygoogle || []).push({}); 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 ofh. He falls a distance of [tex] \frac{h}{n} [/tex] in the last interval of time of∆tof his fall. What is the heighthof 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 ofh. He falls a distance ofh/4in the last interval of time of1.0 secof his fall. What is the heighthof the building?"

in another source and the answer is h = 270m

Thanks in advance :)

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# Homework Help: Spiderman Free fall problem

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