Please, pleeeaze ignore my pervious post. Tevildo is correct. It does work. My excuse? It’s early in the morning.
While everyone else here is obviously smarter about the math here, Why am I the first to ask why you would jump off the 3rd floor balcony in the first place?
Because the big, angry boyfriend just got home. 
I feel like I stepped into a thread where I clearly don’t belong, but I’ll ask anyway: If you fall at 32 feet per second per second, then why don’t you travel 32 feet during the first second, not 16?
I swear I’ve read every post in this thread and I’m still not getting why it’s only 16, not 32.
Because 32 ft/s2 is the acceleration due to gravity. If your speed was a constant 32 ft/s you would travel 32 ft in one second. However, the falling guy starts at 0 ft/s and during that second is accelerated up to 32 ft/s (ignoring air).
Because your speed starts at zero and increases (linearly) to 32 ft/s over that second. After half a second, your speed is only 16 ft/s.
You start out at 0 fps. At the end of one second you are travelling at 32 fps.
The distance you travel HAS to be less that 32 feet.
Look at it this way. Start at 32 feps and DON’T accelerate. The distance travelled is 32 feet. If you start out slower, obviously you are going to travel a shorter distance.
brian
Despite a previous post, it’s much easier to understand if you look at your average velocity. If you start at zero and at the end of one second, you are travelling at 32 feet per second, then your average velocity during that second was 0 + 32 / 2 or 16 feet per second. Thus, by definition, you have travelled 16 feet during that second.
Just to clarify one thing – you aren’t “falling at” 32 ft./s[sup]2[/sup] as that implies a speed (e.g. falling at 120 miles/hr). As has been explained, this is the rate at which your speed changes; your acceleration.
Suppose I have a rocket car. It accelerates at 50 miles/hr per second. Starting at 0 mph, if it speeds up, it won’t travel 50 miles in the first second, but it will be going 50 miles an hour after one second has passed.
Well crap, you’re right. I didn’t think on it carefully enough — and me an ex-physics major too. How embarrassing.
Never mind then.
A-ha. There it is then. I distinctly remember being told as a child that “gravity makes things fall to Earth at 32 feet per second per second” and it never left my brain. But thanks to you dispelling that misinformation, and everyone else’s explanations, it now makes sense. Thanks! 
You do actually need calculus to derive the formulae. But once you have the formulae (whether from deriving them yourself, or from taking the book’s or the teacher’s word for them), you don’t need calculus to use them.
Right. Which is a good thing, for me. No matter how hard I’ve tried or what resources I’ve used, no one’s been able to explain the process of calculus to me in such a way as to make the light come on. I understand the reasons for needing calculus and I understand the principles behind it, more or less. But when it comes to actually doing it, I’m a total wash. I dunno what I’m doing wrong or how I could look at it to make it clear in my mind, but I’ve given up trying and have decided I’m hopeless at this point. Maybe you really can’t teach an old dog new tricks. 
I suppose you’ve been through this exercise but I’ll give it one more chance. Example:
Assume the acceleration is constant = a. The acceleration is the time rate of change of velocity.
acceleration is dv(t)/dt = a.
If we integrate the acceleration we will get the velocity, v(t), because the integral of the derivative of a function such as v(t) is just the function itself.
We are now looking for a function whose time derivative is a.
v(t) = a*t will fill the bill. Why? Well because the time derivative of v(t) is the limit of (vt(t + Dt) - v(t))/Dt as Dt → 0. I used D because I don’t know how to get the Greek capital delta. Dt is a small non zero increment of time.
so the derivative of a*t will be found by evaluating the limit of the following as Dt → 0.
(a*(t + Dt) - at)/Dt = (at + aDt - at)/Dt = a*Dt/Dt = a.
When we take the limit as Dt → 0 this is still a and is the derivative of a*t which is what we wanted.
So the integral of dv(t)/dt = v(t) = a*t.
The same process will yield the distance traveled although it is a bit more complicated if done this way.
There are formal ways to integrate a and get the answer a*t directly. Looking the answer up in a table of integrals is easy. Or using mathematical software is even easier.
Uh…hey!
And the reason I used the “jumping off the balcony” example was because it was windy and my spit wasn’t falling in a straight line and wind resistance was screwing up my calculations.
So there.
Ripped from today’s headlines, a man, and a naked man at that, leaps to his death from the 8th floor of a downtown Washington building. As yet, no one knows what the h*ll it was all about.
Computing the duration of his fall and his speed of impact are left as exercises for the reader.
The edition of the standard Serway college physics text that I had in a class once had a problem that revolved around the old labor song “Drill Ye Tarriers, Drill”, about men doing a blasting job:
The question? “What was Jim Goff’s hourly wage?” Show your work.
Neglecting air resistance:
Let x = height, t = time for descent.
x = ut + 1/2at^2
u = 0 ft/s
x = 1 mile = 5280 ft
a = 32 ft/s^2
5280 = 0 * t + 1/2 * 32 * t^2
t^2 = 5280 / 16 = 330 s^2
t = 18.17 s (to four figures)
Total time of flight = 2t = 36.34 s = 0.01092 hours
Rate of pay = 1 / 0.01092 = $99.09 per hour.
Aren’t you making the (potentially untenable) assumption that there was no elevation change between where he left the ground and where he landed? 
As well as the even less tenable assumption that Goff’s supervisor was honest, and docked him for the amount of time actually spent in the air. The problem as stated in the book included “State all of your assumptions”.