# Acceleration/Fall question

1. Feb 21, 2008

### PhysWalk

1. The problem statement, all variables and given/known data
The question is as follows: An object falls from the top of the roof of a building. The object falls past a 3.5 meter high window somewhere on the building ,in 0.47 seconds. Find the distance between the top of the window and the roof.

2. Relevant equations
V=d/t (Velocity = Distance over Time)
Vf^2=Vi^2 + 2ad (Velocity Final squared = Velocity initial squared + 2 times acceleration times distance)

I believe these are the equations that I should be using.

3. The attempt at a solution

3.5/0.47 = 7.44681 = V

So, Vf = 7.44681, Vi = 0 (assuming the object falls from rest), Acceleration = 9.8m/s/s, distance = X

7.44681^2 = 0^2 + 2 (9.8) (d)
55.455=19.6 (d)
55.455/19.6 = 2.82934
d = 2.82934
d = 2.8 meters.

And that is what I have so far, 2.8 meters from the top of the window, to the roof. Would this be the right approach to the problem? Or have I made a mistake?

2. Feb 21, 2008

### <---

The velocity you took is the average velocity for the fall past the window.
That's not what you need. You want to start by finding the velocity initail just as your object reaches the top of the window, you can do this using the position equation, but don't forget that it's still accelerating.

3. Feb 21, 2008

### blochwave

the equation d=v*t tells you how far something has fallen in a time t while traveling at a velocity of v

Problem: The only time you have a constant velocity is when you're not accelerating. Something that's falling most definitely has an acceleration, which you presumably know since the problem is on earth

To approach this you want to start with the window part

It has a some Viw(Vi at start of window) you don't know, and some Vfw(velocity when it finishes passing the window) you don't know, but you DO know the time(.47 seconds)and you DO know the distance(3.5 meters) and the acceleration

You actually don't care about the Vfw, just the Viw, right(you can actually solve it with either but the problem asks for how far to the top of the window)? If you find Viw, we can zoom back out to the whole building and you know how fast it's going when it reaches the window now. So now you know Vi=0(the roof) and Vf=that speed you just found, Viw, and you still know a, so now you can find the associated distance to go from 0 to Viw

4. Feb 21, 2008

### PhysWalk

Hi, and thanks for the replies, in response to blochwave, is this what you are talking about:

d=Vit+1/2at^2, so 3.5=Vi (0.47) + 1/2 (9.8) (0.47)^2 which = 5.143808511, and that is the Vi for the top of the window. So I can use that with:

Vf^2= Vi^2 + 2ad, so 5.143808511^2 = 0 + 2 (9.8) (d)

So, d= 1.349937041, I think

Last edited: Feb 21, 2008
5. Feb 21, 2008

### blochwave

If you didn't mess up one the calculator(I didn't check the actual numbers)it should be good

Remember that your acceleration is -g usually

It worked out ok because when you start from a higher place and fall to a lower place, d is negative(d=final position-initial!), so you forgot two negatives, which gave you velocity as a positive number, and so on

Well you effectively called the up direction negative and the down direction positive, which isn't BAD, in fact if you did it intentionally(don't lie!) it's probably good, but usually you just do the obvious and call down negative and up positive. It's easier that way when you have projectile motion that goes both up and down

We do usually feel safer working without negatives of course, but the danger is if you keep switching, using down as positive for falling motion, and down as negative for projectile, you're gonna forget once, get some weird negatives, and drive yourself insane.

But if you don't forget you just look clever