What is the Average Acceleration of a Person Falling Off a Mountain?

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In summary, Carlos Ragone's rock anchor gave way and he plummeted 494 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.5 ft deep, estimate his average acceleration as he slowed to a stop (that is while he was impacting the snow). Assume a coordinate system where down is positive. The Attempt at a SolutionI split the region into two parts, one where the final velocity would be the initial for the second region. I have a total distance of 497ft(150.6m) for the first region an initial velocity of 0m/s and a final velocity of 53.9m/
  • #1
icesalmon
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Homework Statement


In some amazing situations, people have survived falling large distances when the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 494 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.5 ft deep, estimate his average acceleration as he slowed to a stop (that is while he was impacting the snow). Assume a coordinate system where down is positive.

The Attempt at a Solution

I split the region into two parts, one where the final velocity would be the initial for the second region. I have a total distance of 497ft(150.6m) for the first region an initial velocity of 0m/s and a final velocity of 53.9m/s where the time to fall over 497ft(150.6 m) is 5.5 seconds. now I have to find, for the second region where the body hits the snow over and falls through the snow for a distance of 3.5ft(1.1m), the average acceleration. So I need VF V0 tf and ti I have initial velocity as my final for the first region because that's where he impacted the snow to be 53.9m/s and my final velocity as 0m/s because he is at rest. The initial time I have as the final time for the first region, which is Ti = 5.5 seconds and I need to find the Tf ( the time it took for him to fall through the snow and I'm a bit stuck here.
 
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  • #2
You can use a kinematic equation which does not contain the parameter t to find the average acceleration. This then allows you, if you want it, to determine the time through the snow.
 
  • #3
so (vf)2 - (vi)2 = 2a(xf - xi)
using that equation I'm getting -1320.55m/s2
that seems too fast...plus the answer is incorrect
 
  • #4
When you say "the answer is incorrect", do you have the expected answer? Or is it an automated system telling you your answer is incorrect?

If the latter, consider:
1. Keep more significant digits in all intermediate calculations
2. Don't round until the very end
3. Round to the correct number of significant figures
4. Make sure that your result is presented in the required units
 
  • #5
converting ft -> meters
497/3.3 = 150.606060606060m
finding final time in first interval ((150.606060606060 meters)/ 9.81m/s2)*2 = t2
tf = 5.41173116943646seconds
vf = 0 + (9.81m/s2)*5.41173116943646seconds
vf = 54.35890827721717 = vi for region two
avg acceleration for this problem I'm using( vf - vi ) / ( tf - ti)
now I'm going to use (vf)2 - (v0)2 = 2a(yf - yi) so I have -1(54.35890827721717m/s)2 = 2a(1.1meters) a = - 1343.132231404959 = - 1343.1m/s2
 
  • #6
my units are correct, I haven't rounded until the end. Either way I round my answer is incorrect on the homework site, I'm not sure if this a conceptual problem (although I believe my setup, diagram and all my equations are correctly used) or rounding errors.
 
  • #7
The original problem statement said "he plummeted 494 feet to land in snow." You've used 497. 3.5 ft is 1.067 m to three decimal places. I think you'll find that your final result should be just a tad larger. Round to two significant figures, since "3.5 ft" has just two.
 
  • #8
icesalmon said:
my units are correct, I haven't rounded until the end. Either way I round my answer is incorrect on the homework site, I'm not sure if this a conceptual problem (although I believe my setup, diagram and all my equations are correctly used) or rounding errors.

The method is correct. You used 497 ft in your calculation, while the question states 494 ft. It is good practice to work symbolically through the problem to obtain a single equation for the answer, from which you can simply substitute numbers in. This helps to reduce errors, particularly rounding errors and dimensional inconsistencies.
 
  • #9
I believe this could be estimated much more simply. Convert 494 ft to meters, then multiply by the gravity constant (9.81 m/s^2) to estimate total acceleration right up to the point of impact.

That value is the amount of deceleration that should be experienced for the remaining distance (between impact and when the body comes to rest). So divide the total acceleration by the remaining post impact distance (also converted to meters) to get the average deceleration

Since we are assuming a coordinate system in which down is positive the deceleration must be represented as a negative value.
 
  • #10
leelou said:
I believe this could be estimated much more simply. Convert 494 ft to meters, then multiply by the gravity constant (9.81 m/s^2) to estimate total acceleration right up to the point of impact.
There is no reason to estimate the total acceleration right up to the point of impact. Neglecting air resistance, it is 9.8 m/s2 or 32 ft/s2. It looks like you are confusing velocity and acceleration.
 
  • #11
icesalmon said:
I haven't rounded until the end.
You divided by 1.1m near the end, to represent 3.5ft. You earlier used a ratio of 3.3. Using that same ratio again would have given a much more accurate answer.

But none of these conversions are necessary. There are many advantages in working purely symbolically, only plugging in numbers at the end. In the present case you could have avoided all reference to velocities. You would get that the free fall drop x g = depth of impact x upward acceleration in snow. Since both heights are in feet, their units cancel, leaving you with 494g/3.5.

I believe this is @leelou's method.
CAF123 said:
You used 497 ft in your calculation, while the question states 494 ft.
Good point. Mostly these questions ask for the force exerted during deceleration, in which case it is appropriate to consider the total descent, 494+3.5ft. But here we are asked for the net acceleration in snow, so that already includes the continued affect of gravity.

Soap box:
The question should not be asking for "average acceleration". It should be telling you to assume constant acceleration in the snow (though in practice it would increase a little with depth).
If the acceleration were to vary substantially, as with SHM say, you would need the duration to compute its average. This is because average acceleration is defined as Δv/Δt, and if acceleration varies then this is not necessarily the same as Δ(v2)/(2Δs).
 
  • #12
kuruman said:
It looks like you are confusing velocity and acceleration.
You may be misunderstanding leelou's method. See my post #11.
 
  • #13
haruspex said:
You may be misunderstanding leelou's method. See my post #11.
I saw only one interpretation in the statement
leelou said:
... to estimate total acceleration right up to the point of impact.
and responded accordingly. I suppose it takes a haruspex to divine what a mere kuruman cannot. :smile:
 

1. How often do people fall off mountains?

The exact frequency of people falling off mountains is difficult to determine, as it can vary depending on factors such as location, type of mountain, and individual experience. However, according to a study published in the American Journal of Emergency Medicine, approximately 10,000 people fall off mountains every year in the United States alone.

2. What are the most common causes of falling off a mountain?

The most common causes of falling off mountains include human error, such as misjudging terrain or taking unnecessary risks, as well as environmental factors like loose rocks, sudden changes in weather, or fatigue. Lack of proper equipment and training can also contribute to falls.

3. What are the most important safety precautions to prevent falling off a mountain?

The most important safety precautions to prevent falling off a mountain include proper training and experience, use of appropriate equipment, being aware of weather conditions, staying on marked trails, and not taking unnecessary risks. It is also crucial to always have a plan in case of an emergency and to inform someone of your intended route and estimated return time.

4. Can falling off a mountain be fatal?

Unfortunately, yes. Falling off a mountain can be fatal, especially if the fall is from a significant height or if the person is not adequately prepared or trained. According to the National Park Service, falls are the leading cause of death in national parks, with an average of 25 fatalities per year.

5. How can someone survive a fall off a mountain?

Surviving a fall off a mountain depends on various factors, such as the height of the fall, the terrain, and the individual's physical condition. However, some ways to increase the chances of survival include having a plan for emergencies, using proper safety equipment such as helmets and ropes, and practicing self-arrest techniques in case of a fall. It is also essential to seek immediate medical attention after a fall, even if there are no apparent injuries.

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