# duh- basic physics. the speed of a falling object in the absence of air? **URL:** \ **Category:** Factual Questions\ **Created:** [June 16, 2006, 9:56pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999 "2006-06-16T21:56:05Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![CC](https://avatars.discourse-cdn.com/v4/letter/c/f1d935/32.png) [@CC](https://boards.straightdope.com/u/CC)\ **Post date:** [June 16, 2006, 9:56pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/1 "2006-06-16T21:56:05Z")
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. ---
**Author:** ![Rhubarb](https://avatars.discourse-cdn.com/v4/letter/r/59ef9b/32.png) [@Rhubarb](https://boards.straightdope.com/u/Rhubarb)\ **Post date:** [June 16, 2006, 10:00pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/2 "2006-06-16T22:00:02Z")
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. 😃 ---
**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\ **Post date:** [June 16, 2006, 10:15pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/3 "2006-06-16T22:15:58Z")
1. Falling where? On another planet? The Earth has air. 2. Just tell the questioner that the object is not falling, it’s the planet that’s moving towards the object. 3. 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. ---
**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\ **Post date:** [June 16, 2006, 10:17pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/4 "2006-06-16T22:17:01Z")
and a = 9.81 m s[sup]-2[/sup] ---
**Author:** ![CC](https://avatars.discourse-cdn.com/v4/letter/c/f1d935/32.png) [@CC](https://boards.straightdope.com/u/CC)\ **Post date:** [June 16, 2006, 10:39pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/5 "2006-06-16T22:39:17Z")
Arnold - two things: 1. I know the Earth has air. It’s a hypothetical question 2. 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. ---
**Author:** ![Lemur866](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lemur866/32/434_2.png) [@Lemur866](https://boards.straightdope.com/u/Lemur866)\ **Post date:** [June 16, 2006, 10:56pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/6 "2006-06-16T22:56:09Z")
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. ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [June 16, 2006, 11:01pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/7 "2006-06-16T23:01:05Z")
There are two ways to derive the formula d=1/2 a t^2. The easy (read: non-calculus) 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. ---
**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\ **Post date:** [June 16, 2006, 11:04pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/8 "2006-06-16T23:04:03Z")
> [@CC](#): > > Arnold - two things: > your formula isn’t helpful. (rending garments) I’m so sorry!!! ---
**Author:** ![Happy\_Clam](https://avatars.discourse-cdn.com/v4/letter/h/3bc359/32.png) [@Happy\_Clam](https://boards.straightdope.com/u/Happy_Clam)\ **Post date:** [June 16, 2006, 11:08pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/9 "2006-06-16T23:08:26Z")
> [@Arnold Winkelried](#): > > (rending garments) I’m so sorry!!! So you should be! We expect a show trial, a confession and immediate hari-kiri on your part. ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [June 16, 2006, 11:13pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/10 "2006-06-16T23:13:06Z")
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. ---
**Author:** ![Lemur866](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lemur866/32/434_2.png) [@Lemur866](https://boards.straightdope.com/u/Lemur866)\ **Post date:** [June 16, 2006, 11:13pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/11 "2006-06-16T23:13:06Z")
Actually, Arnold really gave you everything you need. The only part you were missing was v=at. ---
**Author:** ![Lemur866](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lemur866/32/434_2.png) [@Lemur866](https://boards.straightdope.com/u/Lemur866)\ **Post date:** [June 16, 2006, 11:14pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/12 "2006-06-16T23:14:13Z")
By “terminal velocity”, I assume CC meant “velocity when the object hit the ground after falling 1000 meters”. ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [June 16, 2006, 11:14pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/13 "2006-06-16T23:14:57Z")
But, since the subject of this topic says “in the absence of air”, I can see where the confusion arises. D’oh! ---
**Author:** ![Shagnasty](https://avatars.discourse-cdn.com/v4/letter/s/9dc877/32.png) [@Shagnasty](https://boards.straightdope.com/u/Shagnasty)\ **Post date:** [June 16, 2006, 11:19pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/14 "2006-06-16T23:19:01Z")
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. ---
**Author:** ![Happy\_Clam](https://avatars.discourse-cdn.com/v4/letter/h/3bc359/32.png) [@Happy\_Clam](https://boards.straightdope.com/u/Happy_Clam)\ **Post date:** [June 16, 2006, 11:20pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/15 "2006-06-16T23:20:42Z")
> [@Lemur866](#): > > By “terminal velocity”, I assume CC meant “velocity when the object hit the ground after falling 1000 meters”. 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. ---
**Author:** ![Lemur866](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lemur866/32/434_2.png) [@Lemur866](https://boards.straightdope.com/u/Lemur866)\ **Post date:** [June 16, 2006, 11:28pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/16 "2006-06-16T23:28:29Z")
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. ---
**Author:** ![Happy\_Clam](https://avatars.discourse-cdn.com/v4/letter/h/3bc359/32.png) [@Happy\_Clam](https://boards.straightdope.com/u/Happy_Clam)\ **Post date:** [June 16, 2006, 11:33pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/17 "2006-06-16T23:33:09Z")
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? ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [June 17, 2006, 2:33am UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/18 "2006-06-17T02:33:07Z")
> [@Happy Clam](#): > > 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=m_g_h Kinetic Energy=1/2_m_v[sup]2[/sup] Setting those equal, and solving for v yields: v=[2_g_h][sup]1/2[/sup] Where g=gravity (9.81 m/s[sup]2[/sup] 32.2ft/s[sup]2[/sup]), and h=height ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [June 17, 2006, 6:02am UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/19 "2006-06-17T06:02:29Z")
> [@](#): > > That is true. Another way (easier if you ask me) is to do an energy balance. 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. ---
**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\ **Post date:** [June 17, 2006, 2:33pm UTC](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999/20 "2006-06-17T14:33:29Z")
I wish (gasp) you would have told me that (arrrghhh) soone… [Next page](https://boards.straightdope.com/t/duh-basic-physics-the-speed-of-a-falling-object-in-the-absence-of-air/360999.md?page=2)