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

• Brunno
In summary, the conversation discusses a problem involving an object falling from a height with no initial velocity. One person suggests solving the problem using the formula V = \sqrt{2.g.h}, while another suggests using t = \frac{h}{g}. The first person asks where the second formula came from and the second person explains their thought process. A third person joins the conversation and expresses confusion about the equations being used. The second person provides a link to where they found the resolution to the problem and asks if there is another way to solve it. The third person points out that without showing how the equations were derived, it's difficult to determine if there is another way to solve the problem. The conversation ends with the second person defending their
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.

## The Attempt at a Solution

Brunno said:
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.

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

Okay, it may be like

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

but how does that relate to the kinematic equations?

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.

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?

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:

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

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 said:
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?

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?

#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?

zgozvrm said:
I take that back ... it does not work!

What value of t did you come up with?

Or does it mean what?

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?

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!

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?

## 1. What is free-fall and how does it differ from regular falling?

Free-fall is the motion of an object falling under the influence of gravity without any other external forces acting on it. This means that the object is only affected by the force of gravity, and not by air resistance or other forces. In regular falling, other forces such as air resistance may affect the motion of the object.

## 2. How do you calculate the time of fall in a free-fall problem?

The time of fall can be calculated using the equation t = √(2h/g), where t is the time of fall, h is the initial height of the object, and g is the acceleration due to gravity (usually 9.8 m/s² on Earth). This equation assumes that the object is dropped from rest and there is no air resistance.

## 3. What is the significance of finding the time of fall in a free-fall problem?

Finding the time of fall allows us to determine how long it takes for an object to reach the ground in a free-fall situation. This information can be useful in various applications, such as calculating the velocity of the object at impact or predicting the trajectory of a falling object.

## 4. Can the time of fall be affected by factors other than initial height and gravity?

In a free-fall problem, the time of fall is only affected by the initial height and the acceleration due to gravity. Other factors, such as the mass or shape of the object, do not affect the time of fall.

## 5. Are there any limitations to using the equation t = √(2h/g) to calculate the time of fall?

Yes, this equation assumes that the object is dropped from rest and there is no air resistance. In real-world situations, objects may have an initial velocity or encounter air resistance, which can affect the time of fall. Additionally, this equation only applies to objects falling near the surface of the Earth, as the acceleration due to gravity may vary on other planets or in outer space.

• Introductory Physics Homework Help
Replies
28
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
450
• Introductory Physics Homework Help
Replies
34
Views
897
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
776
• Introductory Physics Homework Help
Replies
25
Views
602
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K