I fell 20ft through the ceiling; what was my force of impact?

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SUMMARY

The discussion centers on calculating the force of impact from a 20-foot fall through a ceiling. The individual, weighing 108.862 kg, fell a distance of 6.096 meters, with a calculated velocity of 10.93 m/s just before impact. The force of impact was estimated using the change in momentum, yielding a force of approximately 4762.207 N (1070.59 lbs) assuming a deceleration time of 0.25 seconds. The conversation highlights the complexities of accurately determining impact force without knowing the specifics of the fall's dynamics.

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  • Familiarity with the equations of motion, particularly free fall calculations.
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This discussion is beneficial for physics students, engineers, safety professionals, and anyone interested in understanding the dynamics of falls and impact forces.

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So I'm trying to figure out a rather amusing problem. A month ago I was doing some work in my attic when I took a wrong step and plunged through my drywall ceiling and fell a straight 20ft to the very bottom of my stairwell landing. I was knocked unconscious instantly and don't remember anything about the fall.

Now that I'm fully recovered I'm curious as to what the force of impact equated to. Was it like being kicked in the head by a mule? A right hook from a pro boxer? Being slapped with a large fish? My problem though is that I have no idea how to calculate the force of impact without knowing how high/far I bounced after landing. The stairwell landing is a rather constricted space; all I know is that I bounced/slumped into the rather solid iron bars of the railing with such force that I bent them out of shape (they're solid enough that I can't bend them back).

Can anyone offer some suggestions that don't involve a crash test dummy and high speed camera? Is there any way to calculate force of impact without the bounce? Here's the data that I was able to figure out:

Mass: 108.862 kg
Distance: 6.096m from floor to ceiling
Time to fall: 1.115s
Velocity: 10.93m/s (24.3 mph)
Kinetic energy before impact: 6502
 
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1. Why should there be a single well-defined "impact force"?

2. Assuming that there were a single well-defined "impact force", what physical significance would it have?

3. I expect that you have retrograde amnesia and do not remember the exact details of the fall. How do we know whether the fall began from a standing position above the drywall, from a prone position on the drywall or from a hanging position below the drywall? How do we know whether the fall ended with a supine impact on the floor below or an upright impact, rolling from the feet onto the back, parachutist style? Those details could alter the distance of free fall from the nominal 6 meters down to 4 meters or up to 7 meters. And they could alter the decelleration distance from a nominal 1 meter down to as low as a few centimeters.

The resulting acceleration could then be as low as 4 g's or as high as 210 g's.

Given the fact of your survival, the likelihood of a figure toward the high end of that range can be considerably reduced.
 
Hi jbriggs, thanks for replying!

1. Why should there be a single well-defined "impact force"?

Given that it was a straight vertical drop onto a flat surface, I figure the force of impact on landing is the important bit. The bounce/slump that sent me face first into the solid iron bars of the railing is a factor too, but I don't know how I can estimate the force of the second impact without knowing the first.

2. Assuming that there were a single well-defined "impact force", what physical significance would it have?

It bounced me through some very solid iron bars face-first. Plus it makes for an amusing anecdote.

3. I expect that you have retrograde amnesia and do not remember the exact details of the fall. How do we know whether the fall began from a standing position above the drywall, from a prone position on the drywall or from a hanging position below the drywall? How do we know whether the fall ended with a supine impact on the floor below or an upright impact, rolling from the feet onto the back, parachutist style? Those details could alter the distance of free fall from the nominal 6 meters down to 4 meters or up to 7 meters. And they could alter the decelleration distance from a nominal 1 meter down to as low as a few centimeters.

I remember what happened just before the fall. I was standing fully upright on a wooden beam in the attic, then stepped off it with both feet (there were thin plywood sheets placed across the attic beams, and it was quite dark, so I couldn't tell what I was standing on. The thin plywood couldn't hold my full weight in the middle so I fell through that and the drywall ceiling). So, the fall started with me standing upright, and my injuries were indicative of a legs-first landing. I then bounced or slumped face first into the railing.
 
It is still a lot of guesswork. Assuming .5m of deceleration length (mainly based on your legs), the average deceleration would be 6m/.5m * 1g = 12g. Nothing you want to happen, but unless you hit anything else in some bad way it is not too dangerous to have this acceleration for ~100ms.
 
Elusiv3Pastry said:
So I'm trying to figure out a rather amusing problem. A month ago I was doing some work in my attic when I took a wrong step and plunged through my drywall ceiling and fell a straight 20ft to the very bottom of my stairwell landing. I was knocked unconscious instantly and don't remember anything about the fall.

Now that I'm fully recovered I'm curious as to what the force of impact equated to. Was it like being kicked in the head by a mule? A right hook from a pro boxer? Being slapped with a large fish? My problem though is that I have no idea how to calculate the force of impact without knowing how high/far I bounced after landing. The stairwell landing is a rather constricted space; all I know is that I bounced/slumped into the rather solid iron bars of the railing with such force that I bent them out of shape (they're solid enough that I can't bend them back).

Can anyone offer some suggestions that don't involve a crash test dummy and high speed camera? Is there any way to calculate force of impact without the bounce? Here's the data that I was able to figure out:

Mass: 108.862 kg
Distance: 6.096m from floor to ceiling
Time to fall: 1.115s
Velocity: 10.93m/s (24.3 mph)
Kinetic energy before impact: 6502

Your velocity right before you hit the floor was

v=√(2gh) = 10.936 m/sec. [assuming a free fall]

your momentum was

p = 1190.552 Kg.m/sec

Now we have to make a guess. The guess is your body came to rest on the floor in .25 sec (.5 sec seems too long).

The change in momentum Δp = 1190.552 Kg.m/sec. and time Δt = .25 sec

Using the Force formula

F = Δp/Δt = 4762.207 N = 1070.59 lbs

Someone please verify the number using a different method.
 
Thank you very much for the help! :)
 
Elusiv3Pastry said:
So I'm trying to figure out a rather amusing problem. A month ago I was doing some work in my attic when I took a wrong step and plunged through my drywall ceiling and fell a straight 20ft to the very bottom of my stairwell landing. I was knocked unconscious instantly and don't remember anything about the fall.

Now that I'm fully recovered I'm curious as to what the force of impact equated to. Was it like being kicked in the head by a mule? A right hook from a pro boxer? Being slapped with a large fish? My problem though is that I have no idea how to calculate the force of impact without knowing how high/far I bounced after landing. The stairwell landing is a rather constricted space; all I know is that I bounced/slumped into the rather solid iron bars of the railing with such force that I bent them out of shape (they're solid enough that I can't bend them back).

Can anyone offer some suggestions that don't involve a crash test dummy and high speed camera? Is there any way to calculate force of impact without the bounce? Here's the data that I was able to figure out:

Mass: 108.862 kg
Distance: 6.096m from floor to ceiling
Time to fall: 1.115s
Velocity: 10.93m/s (24.3 mph)
Kinetic energy before impact: 6502
Given the information that you presented, one can calculate the linear impulse of your fall. The linear impulse is the integral of the force over time. One can not calculate a force alone with the information that you were given.
The impulse is equal to the total change in linear momentum. Therefore,
Δp=mΔv
where Δp is linear momentum, m is mass and Δv is change in velocity. The impulse is:
Δp=108.862 kg(10.93 m/s)
Δp≈1190 kg m/s
Δp≈1190 Newton Seconds

Either units are correct. Many physicists prefer seeing the answer in the second form.
I did not check if your value for velocity is correct.
This answer is not a force, which has units of Newtons. To calculate the force (Newtons) as a function of time (Seconds), more information is needed.
 

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