# Homework Help: Homework Due Tonight (jan 27)

1. Jan 27, 2009

### papa_smurf493

1. The problem statement, all variables and given/known data
Hi im just starting a college physics class, all the work is pretty much online and im am having a terrible time learning all the equations and things of that nature. The question is

You are standing by your living room window when an object falls vertically downward in front of you. You observe that it takes 0.210 seconds for the object to pass the window, which is 1.35 m tall. From what height did the object fall from above the window?

I know that the object is experiencing 9.8 m/s in freefall but i have no clue where to start as far as equations go. Please help im terrible at this.
2. Relevant equations

3. The attempt at a solution

2. Jan 27, 2009

### papa_smurf493

Im sorry that i cant really provide any work, im definately not trying to freeload, and i do want to learn how to do this stuff. I just need someone to walk me through it.

3. Jan 27, 2009

### LowlyPion

Welcome to PF.

You just need to know that the distance an object falls is given by:

x = 1/2*g*t2

Since you know the difference in time and you know the distance it traveled in that additional time then you just subtract one equation for one distance from the other. And since 1 time is longer by the time they give you ... then just solve the resulting quadratic equation.

4. Jan 27, 2009

### papa_smurf493

what is the g in the equation? im guessing x would be the height and t is the time

5. Jan 27, 2009

### fwFAWFSERG

im not too sure, what i would probably work towards would be:
well it took the object .210 seconds to move 1.35 meters, so find out that initial velocity (when it first is visible from the window)

then you can use the velocity you solved for (when the object is at the top of the window/ first visible) and use that as the final velocity again, and solve for the distance.

edit: g is gravity, and lowlys method is good too.

6. Jan 27, 2009

### The Dagda

Last edited: Jan 27, 2009
7. Jan 27, 2009

### papa_smurf493

so my equation should look like this

x=1/2 * 9.8m/s * .210 Squared

8. Jan 27, 2009

### papa_smurf493

V= 9.8m/s Squared *.210 gives me 2.058m/s squared thats using the forumula V=gt is this correct for the initial velocity?

9. Jan 27, 2009

### papa_smurf493

10. Jan 27, 2009

### LowlyPion

No. The .210 is the time difference.

X1 = 1/2*9.8*t12
X2 = 1/2*9.8*t22

And you know that t2 = t1 + .210
And x2 - x1 = 1.35

The rest is algebra.

11. Jan 27, 2009

### papa_smurf493

im still lost sorry i must just be stupid

12. Jan 27, 2009

### papa_smurf493

how do i go about finding what t1 is?

13. Jan 27, 2009

### LowlyPion

Sorry, I won't solve it for you.

It won't do either of us any good.

Just keep trying.

14. Jan 27, 2009

### papa_smurf493

ok then can you explain it in a different way

i have no idead how to find t1

15. Jan 27, 2009

### papa_smurf493

ok heres what im trying to think of in my head. Basically the question states that it takes a object .210seconds to cover 1.35meters. So with that given information i have to find what height the object falls from. Well a given is the objects acceleration is 9.8m/s squared

16. Jan 27, 2009

### papa_smurf493

ok i may be stupid or something, this is suppose to be easy and i ve been working on it for like a hour please help

17. Jan 27, 2009

### papa_smurf493

ok i cam up with this. Will this equation work?

D=Vi *t +1/2 * a * t2

0*.210+1/2*9.8*?

18. Jan 27, 2009

### papa_smurf493

am i that lost that nobody can help me?

19. Jan 27, 2009

### papa_smurf493

i guess it would help if i found the avg speed which is 1.35/.210 = 6.43m/s

20. Jan 27, 2009

### papa_smurf493

AVG Speed = 1.35/.210 = 6.43 m/s
a = (vm - vt)/t
9.81 = (6.43 - vt)/(0.105)

vt = 5.40 m/s

5.402 = 0 2 + 2(9.81)d

d = 1.49 m
Object was released 1.49 m above window.

is this correct?

21. Jan 27, 2009

### papa_smurf493

wow this is a great forum, i really appreciate all the replys im getting and the help! You guys are great!

22. Jan 27, 2009

### LowlyPion

23. Jan 27, 2009

### papa_smurf493

well you could have broke it down a little better and explained what i needed to do instead of thowing a formula at me and sayin use it

24. Jan 27, 2009

### Kaizer6

Any basic mechanics book solves this problem within the first couple of chapters, take a book outfrom the library on classical mechanics and do some reading. You'll understand what youre trying to do then.

25. Jan 28, 2009

### The Dagda

This book is excellent and free and covers all you need to know about the basics, Courtesy of Motionmountain via Greg Bernhardt.

http://www.motionmountain.net/

I downloaded it yesterday as I'm a little rusty on this stuff, and it is perfect for pre-degree studies.

$$z(t)=z_0+v_0(t-t_0)-\frac{1}{2}g(t-t_0)^2$$

z(t) = height
z_0 = initial height
t_0 = time fall starts
g = 9.8m/s2

Probably too late for this to help but meh. For more information check out page 74 of that text.

Last edited: Jan 28, 2009