High fall on soft surface or lower fall on hard surface?

March 29, 2005 8:46 AM

Basic physics. Sitting on a 2,5m (8.2ft) high boulder without shock-absorbing footwear, should I prefer jumping down 2m (6.6ft) onto a flat rock or the full 2,5m onto a softer forest surface?

I'm wearing climbing shoes: 2mm (0.08") of rubber, have no crashpad to land on and weigh 66kg (145pound). It's getting dark.

posted by Akeem to sports, hobbies, & recreation (14 comments total)

f = m x a
a = deltav/deltat
When you jump on the hard surface, your deltav is lower because you start lower, but your deltat is smaller. The smaller the deltat the greater the acceleration. The greater the acceleration, the greater the force.

When you jump on the soft surface, your deltav is higher because you start higher, but your deltat is larger. The larger the deltat, the smaller the acceleration. The smaller the acceleration, the smaller the force.

Without knowing the deltat, you can't say which is greater.

If you want to experience the power of deltat, get up on a chair and step off it landing by bending your knees a great deal as your hit. Then try it again and keep your knees locked. Big difference.
posted by plinth at 8:59 AM on March 29, 2005

I would downclimb a tree or the boulder rather than jump 6 feet. Not a physics answer, but, that's the way I ussually try to go when bouldering.
posted by trbrts at 9:06 AM on March 29, 2005

plinth, wouldn't the acceleration in both cases be the same? In both cases you would measure the acceleration at 9.8 m/s^2, i.e. the acceleration due to gravity.
posted by mfbridges at 9:18 AM on March 29, 2005

m*g*h = (1/2)*m*V^2 (this is conservation of energy)

Therefore, Velocity = sqrt(2*g*h), where h is the height from which you are jumping, and g is the constant acceleration of gravity (9.8 m/s^2) [as mfbridges says].

Since your final velocity goes as the square root of your height, a 25% increase in height will only increade your landing velocity by around 11%, suggesting that it's worth going for the softer surface.

Also: if I'm remembering my climbing days accurately, I would not want to jump onto rock with climbing shoes at all. Not only is there little padding or protection from jagged edges, but the low traction would make me afraid of slipping upon landing. I guess it depends on the boulder though: one that's fairly smooth, flat, and grippy, wouldn't be so bad.
posted by kickingtheground at 9:30 AM on March 29, 2005

mfbridges: plinth is talking about the deceleration that occurs when the ground is impacted. As he points out, it depends on the impact velocity and the amount of cushioning (the delta_t) on the surface.
posted by muddgirl at 9:31 AM on March 29, 2005

actually you're measuring the decelleration when you land going a certain speed.

you're going faster after falling farther, but the decelleration is less on a softer surface.

somebody better hurry and give this guy a solid answer before he peels off! a controlled fall in either case is probably better than an unplanned one.

my gut feeling is go for the forest floor and roll as you land
posted by jacobsee at 9:31 AM on March 29, 2005

Potential Energy = mass * g * height

For lower jump:

PE = 70 * 9.8 * 2 (for 70 kilogram person)
PE = 1372 Joules

For higher jump:

PE = 70 * 9.8 * 2.5
PE = 1715 Joules

If the ground absorbs more than 343 joules (1715-1372), your body will have to absorb less energy jumping on to the ground than jumping onto the rock. I have no idea how to measure that.
posted by mfbridges at 9:33 AM on March 29, 2005

Oops...i'm wrong.
posted by mfbridges at 9:33 AM on March 29, 2005

Is this really a physics problem? Without knowing the impact absorption of the forest floor, formulae don't really assist much do they? I mean, the acceleration is gravity and there's an extra 1/2 metre so a bit of an increase in velocity.

What sort of forest floor seems the next logical question followed by just how physically fit and flexible are you?

I think you would prefer the known - you are relatively light and I'm assuming you are fit/young/flexible - 2m is easily do-able but 2.5m drop might be risky - unseen uneveness such as branches/rocks/potholes may cause injury.
posted by peacay at 12:04 PM on March 29, 2005

I would hardly call a 2m fall "doable", as it is greater than my own height. Generally, a 6ft fall is considered great enough to do some damage to legs, limbs, or your head.

However, if you can manage to hang from the rock by your fingertips, you could decrease the falling distance considerably, though you lose the advantage of bending your knees.

I think the best bet in this situation is to climb or slide down the side of the rock. Friction will slow your fall, though it might rough up your clothing.
posted by muddgirl at 12:37 PM on March 29, 2005

My assumptions are: Akeem is a guy which probably means that he's physically capable of a 2m drop and that this is a conundrum in so far as he only has the 2 choices that he outlines with no alternative mitigating possibilities such as shimmying his way down.
And 2m is better than 2.5m. Just my take.
posted by peacay at 1:30 PM on March 29, 2005

I'm going to give you a framework for thinking about the problem, then a rough guesstimate for an answer. Sorry this is so long, but this is a hard question.

What you're asking for is the Energy Absorbance of your landing material.

As other have mentioned, your impact energy is equal to your original potential energy: m⋅g⋅h.
Your falls are (rounded to the tens):
E_soil = 1620 J
E_rock = 1390 J
But it turns out that the ratio of heights is all that matters: E_rock is 80% of E_soil.

What's really important is termed impact energy absorbance (or energy dissipation, it has many names), the fraction of how much energy is dissipated by your landing material. For any fall a certain fraction of the landing energy is dissipated (turned into plastic deformation, heat and sound) by the impact material. I'm going to call the amount of energy absorbed by the landing surface divided by the total amount of impact energy α for absorbance. For the two falls, I'll use α_soil for the longer drop onto soil and α_rock for the shorter drop onto the rock. I'm ignoring your shoes to make this simple; they matter a bit for the shorter fall onto rock, but probably don't onto the soil.

The amount of energy your knees and ankles have to absorb is (1-α) ⋅ mgh. This is the quantity you want to minimize.

To your question, it's handy to think of the two of absorbed energy for both falls:
α_soil ⋅ E_soil and α_soil ⋅ E_rock

Remember that E_rock is 80% of E_soil.

So, if α_soil is more than 80% of α_rock, then falling onto the soil is going to be harder on your knees than the rock. If it's less, then you want to land on the soil. That's our framework.

Now comes the messy and complicated bit that I've conveniently swept under the rug: what are the values of α_rock and α_soil? As it turns out, this is a very hard question to answer. Energy dissipation, α, is dependent on many factors: impact speed, the size and shape of the impacting body (i.e. you), the landing material, water content, temperature, and so on.

The first we can make a stab at: rock is a very poor absorber of energy in this domain. The collision will be mostly elastic; α_rock is very close to 0. Your body will have to absorb most of that 1390 J.

The second is the hard one. Whole textbooks are written on the compressibility of soils, but generally soils are good energy absorbers, forest floor duff particularly. Soils nominally have α's close to 1, but that's over the lifetime of the collision. Peak impacts on soils can still be pretty significant. For that small a difference, 80%, however, I'd say that it's a pretty safe bet to fall onto the soil, all things being equal.

For view of the problem, look here. I've taken a very simple bird's eye view here. The authors of that paper go into more detail and consider peak accelerations on impact which is the major contributor to how "hard" a fall feels.
posted by bonehead at 2:07 PM on March 29, 2005

you can guesstimate the peak impact by looking at the distance travelled during decelleration. it's going to be a few mm on a rock, due to sole flexure, and maybe 1-2cm on soil. if you assume constant deceleration then it's a factor of around 10 lower on soil which more than makes up for the extra energy. BUT that assumes you fall with rigid body, i guess, or rather, is only giving you an idea of how much your feet will "sting". i think the best comment above is about rolling. you want to be landing feet together and rolling over, like a parachute landing, and that's going to hurt on rock.
posted by andrew cooke at 4:48 PM on March 29, 2005

I waited for a rabbit to pass by and jumped on that to break my fall.
posted by Akeem at 1:49 AM on November 7, 2005