Free Fall Calculation: Accurate Distance in Seconds

In summary, the app I was using was not accurate for calculating the distance an object falls in a given amount of time, but the formula provided by Doc Al is accurate.
  • #1
icon kid
2
0
I'm a 3D artist and I'm making an animation that involves a free falling object. I know that, in a vacuum at least, an object will fall 32 feet during the first second; during the second second it'll fall 64 feet -- a total of 96 feet after only two seconds of free falling. The problem is that I need to know exactly how far an object has fallen at precise fractions of a second which is difficult to figure out unless someone knows of an easy, handy formula.

Each second of time is divided into 30 parts (frames). In this case I'm "keying" every other frame so the chart for this would look like the one below. In Frame 0 the object is at it's holding position but then it's let go to start falling. In Frame 2, the object has been falling for only a 15th of a second (0.0666 seconds) but I need to know exactly how far it traveled because I have to manually place the object at that point in space at that instant of time and then reposition it again and again after each 15th of a second passes, -- in order for the acceleration to be accurate and look natural.


-- 30 frames/second --

Frame 0 -- 0 seconds Distance = 0 feet
Frame 2 -- 1/15th second (0.0666) Distance = ? feet
Frame 4 -- 2/15th second (0.1332) Distance = ? feet
Frame 6 -- 3/15th second (0.1998) Distance = ? feet
Frame 8 -- 4/15th second (0.2664) Distance = ? feet
Frame 10 -- 5/15th second (0.3333) Distance = ? feet
Frame 12 -- 6/15th second (0.3996) Distance = ? feet
Frame 14 -- 7/15th second (0.4662) Distance = ? feet
Frame 16 -- 8/15th second (0.5328) Distance = ? feet
Frame 18 -- 9/15th second (0.5994) Distance = ? feet
Frame 20 -- 10/15th second (0.6666) Distance = ? feet
Frame 22 -- 11/15th second (0.7326) Distance = ? feet
Frame 24 -- 12/15th second (0.7992) Distance = ? feet
Frame 26 -- 13/15th second (0.8658) Distance = ? feet
Frame 28 -- 14/15th second (0.9324) Distance = ? feet
Frame 30 -- 1 second (1.0000) Distance = 32 feet

I found a free app that does free fall calculations but something seems wrong with the output I'm getting from it.

Here's a screen shot it.
http://www.jeffs-icons.com/free_fall_calc.gif

I'm using the calculation method for objects falling in a vacuum so the size (volume) and weight (mass) of the object as well as atmospheric parameters are irrelevant. In this case I'm solving for the distance fallen in a given amount of time. Time is expressed as numerical values only, and all values are interpreted as seconds, so a duration of "1 minute and 10 seconds" has to be entered as simply 70. To test the accuracy of this app before using its output, I entered a free fall time of one second. It should have given me a distance of roughly 32 feet for a one-second free fall. Instead it gave me 16.077 feet -- and that's far from the mark, almost half.

If you want to try this app yourself go to:
http://www.softpedia.com/get/Science-CAD/Free-Fall-Calculator.shtml
-- click the Free-Fall Calculator 1.0 download button
-- on the next page, right-click the "External Mirror 1" ZIP file and "Save target as." This will download a file called "Free-Fall.zip."

Does anyone know of either an online utility or a downloadable app that will give me the results I expect?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
icon kid said:
I'm a 3D artist and I'm making an animation that involves a free falling object. I know that, in a vacuum at least, an object will fall 32 feet during the first second; during the second second it'll fall 64 feet -- a total of 96 feet after only two seconds of free falling.
This is incorrect. After 1 second the object has fallen 16 feet; After 2 seconds it has fallen a total of 64 feet.

The problem is that I need to know exactly how far an object has fallen at precise fractions of a second which is difficult to figure out unless someone knows of an easy, handy formula.
The formula you need is (where time is in seconds and distance is in feet):
[tex]D = 1/2 g t^2 = 16 t^2[/tex]

More accurately, the acceleration due to gravity (g) is closer to 32.15 ft/s^2 than 32 ft/s^2:
[tex]D = 1/2 g t^2 = 16.075 t^2[/tex]


I found a free app that does free fall calculations but something seems wrong with the output I'm getting from it.
The problem is not with the app, but with your understanding of free fall. Read this: Freefall
 
Last edited:
  • #3
So I've misinterpreted the meaning of "32 feet per second squared" and the output of this free-fall calculator is reliable -- and easier to use than hand-working the formula which you graciously provided.
Thank you Doc Al.
 

1. What is free fall calculation?

Free fall calculation is a mathematical process used to determine the distance an object has fallen due to the force of gravity. It takes into account the acceleration of gravity, which is approximately 9.8 meters per second squared, and the time that the object has been falling.

2. How accurate are free fall calculations?

Free fall calculations are generally very accurate, as long as all the variables such as air resistance and initial velocity are taken into consideration. However, small errors can occur due to rounding or other factors.

3. Can free fall calculations be used for any object?

Free fall calculations can be used for any object that is falling due to the force of gravity, regardless of its mass or shape. However, they may not be accurate for objects that experience significant air resistance or other external forces.

4. How do you calculate the distance of a free fall in seconds?

To calculate the distance of a free fall in seconds, you can use the formula d = ½ * g * t², where d is the distance, g is the acceleration of gravity, and t is the time in seconds. Plug in the values and solve for d to get the distance in meters.

5. Are there any limitations to free fall calculations?

While free fall calculations are generally accurate, they do have some limitations. They only work for objects falling near the surface of the Earth, and they do not take into account external factors such as wind or air resistance. Additionally, they assume a constant acceleration of gravity, which may not be the case in certain situations.

Similar threads

Replies
13
Views
1K
  • Mechanics
Replies
1
Views
1K
Replies
8
Views
1K
  • Science Fiction and Fantasy Media
Replies
2
Views
3K
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
2K
Replies
4
Views
5K
  • Special and General Relativity
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
993
Back
Top