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Homework Help: Person falling off mountain

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data
    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.







    3. 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/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.
     
    Last edited: Sep 4, 2013
  2. jcsd
  3. Sep 4, 2013 #2

    CAF123

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    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.
     
  4. Sep 4, 2013 #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
     
  5. Sep 4, 2013 #4

    gneill

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    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
     
  6. Sep 4, 2013 #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
     
  7. Sep 4, 2013 #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.
     
  8. Sep 4, 2013 #7

    gneill

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    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.
     
  9. Sep 4, 2013 #8

    CAF123

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    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.
     
  10. Jan 21, 2018 #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.
     
  11. Jan 21, 2018 #10

    kuruman

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    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.
     
  12. Jan 21, 2018 #11

    haruspex

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    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.
    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).
     
  13. Jan 21, 2018 #12

    haruspex

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    You may be misunderstanding leelou's method. See my post #11.
     
  14. Jan 21, 2018 #13

    kuruman

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    I saw only one interpretation in the statement
    and responded accordingly. I suppose it takes a haruspex to divine what a mere kuruman cannot. :smile:
     
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