Solving Free-Fall Problems: Find the Time of Fall!

[Brunno]

Homework Statement

An object falls into free falling from a hight such that during the last second of falling it traverses 1/4 of the total hight.Calculate the time of the fall knowing that there's no initial velocity.

Homework Equations

I know solve this problem fallowing these steps:

V = \sqrt{2.g.h}

t = \frac{h}{g}

I'd like to know other(s) way(s) to solve this problem.

Thanks in advanced!

The Attempt at a Solution

 

[zgozvrm]

I know solve this problem fallowing these steps:

V = \sqrt{2.g.h}

t = \frac{h}{g}

Where did you come up with t = \frac{h}{g} ?


It so happens that it works with this problem, but that formula won't work, in general.

 

[Brunno]

Is like Delta S/Delta(a) where a is the accelaration that is equal to g.

 

[zgozvrm]

Okay, it may be like

\frac{\Delta S}{\Delta a}

but how does that relate to the kinematic equations?

 

[Brunno]

You folow the sequence:

Last second:

1 = \frac{\sqrt{2gh - 2g3/4h}}{g}

3/4 og the hight.the total hight minus 3/4h is equal to the 1/4h.

i'm already lost.See if you can go on with this,otherwise i put the rest.That was not an ansewr found by me,But i could understand it and so i wanted to know if there's other way to get to the same answer.

 

[zgozvrm]

How you get from

1 = \frac{\sqrt{2gh - 2g3/4h}}{g}

to

t = \frac{h}{g}

is beyond me...




... and where does

1 = \frac{\sqrt{2gh - 2g3/4h}}{g}

come from anyway?

 

[Brunno]

Let see (seems like I'm the only one that own the solution)

V² = Vo² + 2gh so V = \sqrt{2gh}

g = V/t so t = V/g

If we consider the total hight and subtract 3/4 of it we get to 1/4 of the hight (1/4h).As we see that the last in second :

t = V/g so 1 =\frac{\sqrt{2gh}\sqrt{2g(3/4)h}}{g}

simplifying this we can get to:

\sqrt{h} = \frac{g}{\sqrt{2g} - \sqrt{(3/2)g}}

we see also that :

V = \sqrt{2gh} so \sqrt{2g} . \frac{\sqrt{g}}{\sqrt{2} - \sqrt{3/2}}

Is obvious now as well that t = \frac{V}{g} so just dividing the last equation by g we get to :

\frac{\sqrt{2}}{\sqrt{2} - \sqrt{3/2}} which is already the answer.

At last.


So?

 

[zgozvrm]

It so happens that it works with this problem, but that formula won't work, in general.

I take that back ... it does not work!


What value of t did you come up with?
What about H?

 

[Brunno]

I guess I'll have to pass you the link where i found the resolution.And you will definitely take back what you said.Here it goes:http://www.tutorbrasil.com.br/forum/viewtopic.php?f=9&t=1849


So?

 

[zgozvrm]

#1) I understand the problem and how to answer it; I was making sure you did too. (It wasn't clear from your posts).

#2) What, exactly, did I say that you think I should take back?

 

[Brunno]

I take that back ... it does not work!


What value of t did you come up with?
What about H?

Or does it mean what?

 

[Brunno]

Anyways,we are not paying attention at the reason why i made this thread:Is there another way to solve this problem?If there is which is?

 

[zgozvrm]

Basically, all you've done is give some equations without showing how you derived them (using the kinematic equations as a starting point).
Without knowing what you did to get there, how can we tell you if there is another way to solve the problem?For instance if I were to ask you to solve for x in the following equation

2^x - 4x = 12

and you told me the answer was 5, but wanted to know if there were any other way to solve this equation, I couldn't tell you because I don't know how you solved it in the first place.

Although you have the correct answer, it's not clear how you got it by what you've shown us.By the way, the answer can be simplified:

\frac{\sqrt2}{\sqrt2-\sqrt{3/2}} = 4 + 2\sqrt3
You need to show all your work if you want an answer!

 

[Brunno]

All the work?Come on!I did it.I put everything i know about the question.Let pretend i haven't put the solution and just reading the question you could not find a solution for it?
You mean the question is not complete?Or what?Because as far as i can see the way I did show you on how solve this problem is correctly.I'd like to someone else to see it and say whatthey think.Come on!Is that question that complicated?