Sorry  I’m so removed from this stuff. An object falling one kilometer with no air. How do I calculate its terminal velocity or speed? I know it accelerates at about 10 meters per second per second (ok, 9.8). Damn, I feel so stupid. There’s something so basic that I’m overlooking. d=at^2 ? This stuff is chapter 1.
Simple. s (or d) = 1/2 at[sup]2[/sup]. Thus t = sqrt(2s/a). V=at, so v= a(sqrt(2s/a).
I’m pretty sure that’s right, but someone else will probably come along in about 15 seconds to correct me.

Falling where? On another planet? The Earth has air.

Just tell the questioner that the object is not falling, it’s the planet that’s moving towards the object.

Remember the following formulae:
V = a * t
X = .5 * a * t[sup]2[/sup]
where a is the acceleration, V is the velocity, and X is the displacement from an initial location.
and a = 9.81 m s[sup]2[/sup]
Arnold  two things:
 I know the Earth has air. It’s a hypothetical question
 your formula isn’t helpful. I don’t know how long the fall takes. I only know the distance. The poster above is more helpful.
d = .5 * a * t2
d is distance
a is acceleration
t is time
You know the distance. You know the acceleration. So you can solve for time.
V = a * t
V is velocity, that is, speed.
a is acceleration again.
t is time.
So, you know the acceleration. You know the time, from the first equation. So you can solve for velocity.
There are two ways to derive the formula d=1/2 a t^2. The easy (read: noncalculus) way is this:
The initial velocity is zero, since we are dropping things, not throwing them. The final velocity is v=a t. The AVERAGE velocity is then (0+at)/2 = 1/2 a t, and the total distance covered is (avg. v)*t = 1/2 a t t, or 1/2 a t^2.
Hope that helps. Derivation is always preferable to memorization in my book.
(rending garments) I’m so sorry!!!
So you should be! We expect a show trial, a confession and immediate harikiri on your part.
The issue with air is that air resistance will slow the object down. The formulas given here assume no air resistance. The term “terminal velocity” is used to denote the velocity at which the force of the air resistance on the object is equal to the force of gravity on it. The air resistance, in turn is dependant on the density and shape of the falling object.
Actually, Arnold really gave you everything you need.
The only part you were missing was v=at.
By “terminal velocity”, I assume CC meant “velocity when the object hit the ground after falling 1000 meters”.
But, since the subject of this topic says “in the absence of air”, I can see where the confusion arises.
D’oh!
I just did it in Excel in about a minute.
Start with a cell and keep adding 9.8 to each row (a second) until the sum of the previous rows hits 1000 and take the velocity from that row. I suck at higher math but I am good at getting results anyway.
I don’t think so I was always taught that an object’s terminal velocity is the maximum velocity it can achieve when falling in an atmosphere of a given density, when its acceleration is 0 because the air resistance on the object is equal in force to the pull of gravity.
Sure, but of course if there’s no air there’s no such thing as a terminal velocity.
Since CC explicitly excluded air, I think he just accidentally used the wrong term, and meant to say “final velocity”. Of course, the final velocity is 0. But it’s pretty clear he wanted to know how fast he would be going as he hit after falling 1 kilometer on an airless world with 1 g gravity.
Oh, right, sorry.
Is the 1 kg part important? Since mass has no effect on the amount of force exerted on an object by gravity, in a vacuum it surely doesn’t matter whether the weight is 1kg or 1000kg when it hits the ground, just how long it’s been falling?
That is true. Another way (easier if you ask me) is to do an energy balance. Since there is no energy lost to air resistance, all of the potential energy of the mass goes into velocity. So:
Potential Energy=mgh
Kinetic Energy=1/2mv[sup]2[/sup]
Setting those equal, and solving for v yields:
v=[2gh][sup]1/2[/sup]
Where g=gravity (9.81 m/s[sup]2[/sup] 32.2ft/s[sup]2[/sup]), and h=height
Sure, but that’s not really an explanation, without the calculus or reasoning about the average speed, since now you have to explain why the energy formulae have the form they do.
And for the record, Noodles did include the distance formula in his answer. So no seppuku required.
I wish (gasp) you would have told me that (arrrghhh) soone…