Spiderman Free fall problem

[me_duele_cabeza]

EDIT: Thanks Hootenanny and Doc Al for looking and trying the problem. I figured out my mistake $$h_2 = h - \frac{h}{n}$$ not $$h_2 = 1- \frac{h}{n}$$

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 $$\frac{h}{n}$$ 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 a lot of error):

I made $$h_1 = \frac{h}{n}$$ and $$h_2 = 1- \frac{h}{n}$$

$$V_{final}^2 = 2gh_2$$
$$V_{final} = \sqrt{2gh_2}$$

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

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

I set $$\beta = 0.25g^2\Delta t^2$$ and plug in the values for h1 and h2

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

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

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 :)

 

[Hootenanny]

Try using $$v^2 = u^2 + 2as$$
where $v = final velocity, u = initial velocity, a = acceleration = g, s = displacement = h$
using your working above you could sub $$v^2 = 2gh$$ into the above equation.
Edit: Just a note to say that this solution neglects air resistance.

 

[me_duele_cabeza]

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.

 

[Doc Al]

Try using the kinematic formula relating distance and time:
$$s = 1/2 a t^2$$

 

[me_duele_cabeza]

Thanks for the suggestion. I actually just figured out what I did wrong ( $$h_2 = h - \frac{h}{n}$$ not $$h_2 = 1 - \frac{h}{n}$$). :P

 

[ultimatedisc2]

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

 

[ultimatedisc2]

Scratch that, I was able to rearrange it and understand.

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