# The Physics of Freefall

Without an atmosphere we would continue to accelerate during free fall to ever increasing velocities until we impacted mother earth. Without an atmosphere our parachute would of course be worthless. Hence a soft landing on the moon requires retro rockets to decelerate to a soft landing while parachutes have been used to help decelerate the Martian landers in the thin carbon dioxide atmosphere of mars.

In the absence of atmospheric drag we would experience a linear increase in velocity with time as described by:

Where ln is the natural logarithm base e and cosh is the hyperbolic cosine function.

We can now evaluate eqns (10), (11) and (12) for various times over the free fall period to obtain the acceleration, downward velocity and the distance the skydiver falls. These results are tabulated in Table (1) and corresponding plots are illustrated in Figs (1) through (3).

Eqn (11) was used to calculate the plot in Fig (1). We note that as we exit the aircraft at t = 0, our initial acceleration is 32 ft/sec^2, (gravity rules). As the opposing aerodynamic drag force increases with our increasing free fall velocity, our downward acceleration decreases. We see from Fig (1) that our acceleration diminishes to about half of it’s initial value after 5 sec of free fall and all perceptible downward acceleration has ceased after 15 or 20 sec.

Our free fall velocity was calculated from eqn (10) and is plotted in Fig (2). It steadily increases over the first 5 seconds of free fall from zero to nearly 90 mph. During the next 5 to 10 seconds our acceleration diminishes significantly as we approach terminal. It is the post 10 sec period of the skydive when our sensation of falling is replaced by the feeling being suspended and cradled by the pressure of the wind.

Eqn (12) was use to calculate and plot the free fall distance. It is apparent from Fig (3) that we fall only about 350 feet in the first 5 seconds and at least twice that far in the second 5 seconds.

Beyond 10 seconds the plot is nearly linear as we approach a constant terminal velocity. Fig (3) confirms our often used rule of thumb “we free fall about 1000 feet in the first 10 sec and another 1000 feet for every 5 sec thereafter”. Comparing the distance at 25 sec with that at 20 sec in Table (1) we see a difference of about 860 ft, a bit less than the rule of thumb value of 1000 ft. The 1000 ft per 5 sec of free fall at terminal is only precise for a free fall rate of 1000 ft / 5 sec = 200 ft/sec or 136 mph rather than 120 mph used in this example.

Hopefully this example and discussion may provide some insight to those who are mathematically inclined and curious about the “whys”.

## 12 Comments

**DJL**2014-01-13

Metric, please. These units are only understood by 5% of the world's population.

**virgin-burner**2014-01-13

all those physics and no metric system used - i'm a little disappointed, to say the least!

**erdnarob**2014-01-13

Great work. However there is no clear explanation where the hyperbolic functions come from. Could you make it more explicit please.

**unkulunkulu**2014-01-13

Cool, a good observation that you only need the terminal velocity to calculate everything, didn't think about that!

However, there's one small addition: the starting velocity is not zero, it's horizontal and it does have some impact. You cannot just project everything on the vertical axis, because the drag is quadratic wrt to velocity and you'll accelerate to terminal a bit longer than what this equation implies.

**Daniel11**2014-01-14

this is america we don't is metric system kooks!

**DrDom**2014-01-15

Although I generally apply the metric system in the realm of the sciences, it can be done and easily converted if needed. This is an excellent post and the physics behind freefall is elegant to say the least. log base e and cosh is the hyperbolic function I believe.

I digress though, this is one exceptional post because I'm such a physics nerd and this makes me ponder the sky in ways other than "jump, fall, pull, flare, repeat". Brilliant Brilliant Brilliant. Thank you :)

**Faraway**2014-01-15

Dudes, if you're able to understand those functions, convert ft to m is a piece of cake.

Thanks to Giles for the great explanations!

**Waldschrat**2014-01-15

Metric!! What are in hell is ftÂ˛ ?

**Herckydude**2014-01-15

I bet that a lot more than 5% of the people in skydiving and aviation use the imperial system over the metric system. When I speak for Belgium: Flemish 60% uses imperial, 40% French uses metric in general.

It's not so difficult to convert the given data in this article to the units that are desired by the reader...

**skydivezimbabwe**2014-01-25

This article, and the fact that I'm reading it, just goes to prove that it's always raining somewhere in the world. Since Cessna are produced in USA, we use feet/miles on aircraft instrumentation and flight calculations. However, we use metric units on our charts.

**dplw**2015-11-10

Hi Giles.

Thank you for your first-rate article.

Looking at equations 11 and 12 though, it looks as if a couple of typographical errors may have crept in.

I reckon that equation 11 should be

32*sech^2(0.182*t)

[where "sech^2" represents 'sech squared']

and that equation 12 should be

967*ln(cosh(0.182*t))

[where "ln" represents 'log to the base e']

Kind regards

David Williams

**gon**2016-01-13

Great article Giles.

But for an engineer like myself ....how crazy is to use the imperial units. They make no sense, and complicate all the math. It's funny to see how you mess with the equations to insert conversion factors, like from ft/sec to mph.

It's not that it is impossible to understand, anybody can do the conversions. It just feels weird.

I can understand the tradition and the people using the imperial system in they everyday life. But for engineering and technical stuff... You guys should switch to metric. You see how much easier everything is.

Back to the subjet. Thanks for the article, really good. It reminds me of when I studied all this in the university.

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