# Homework Due Tonight (jan 27)

## Homework Statement

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.

## The Attempt at a Solution

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.

LowlyPion
Homework Helper
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.

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

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.

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so my equation should look like this

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

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?

LowlyPion
Homework Helper
so my equation should look like this

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

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.

im still lost sorry i must just be stupid

how do i go about finding what t1 is?

LowlyPion
Homework Helper
Sorry, I won't solve it for you.

It won't do either of us any good.

Just keep trying.

ok then can you explain it in a different way

i have no idead how to find t1

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

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

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

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

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

am i that lost that nobody can help me?

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

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?

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

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

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.

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.

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