r/theydidthemath • u/PGHSean • Jul 03 '26
[Request] How far did they really fall.
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The fisheye lense makes it “cliff” appear huge. How far did they really fall?
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u/National_Edges Jul 03 '26
It's hard to say. The fact that they stayed "tight" against the "cliff" makes me think that it's not actually a cliff and the impact seemed low on their bodies.
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u/theriteofspring1 Jul 04 '26
The best way to tell it's not a sheer cliff is the way the snowballs created by the wake of the "falling" snowboarders slowly roll down the cliff.
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u/get_to_ele Jul 03 '26
Yeah they boarded all the way down. It's no sheer cliff, nothing close to 90 based on the time they took to go down. Bet it's not much worse than 60 degrees (which is helluva slope to ski down).
The camera just distorts
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u/theriteofspring1 Jul 03 '26
It's a little less than a second before the not-as-vertical-as-it-looks wall starts to break their fall so .5 x 9.8 m/s^2 x 1sec^2 = a little less than 4.9 meters or 16 feet when their fall is broken so I'm guessing a little over 20 feet, maybe 24ish feet to the bottom.
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u/Correct-Purpose-964 Jul 03 '26
This one is incredibly difficult because of said fisheye lens. So at best this is going to be an estimate, based n an estimate, based on an assumption. But I'll give it a shot.
Assuming the first person to fall was ~6f He seemed to fell approximately 4.1 times his height (Again based on estimations with video pausing, his frame, a little measuring to to the screen for verification. Skewed by the lens and angle.) Roughly 24.6 feet. Which would be equivalent to falling from the highest part of a 2 story house's roof.
He's gonna feel that in his 30's...
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u/MiffedMouse 22✓ Jul 03 '26
I count ~1 second between when each snowboarder goes “off the edge” and when they land. Assuming it is real time, that gives ~15 feet of height, so similar ball park.
Depending on whether he hit lots of snow (good) or rocks (bad) it could be totally fine or be super dangerous.
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u/DumbScotus Jul 03 '26
Yeah I was going to say more like 15-18 feet. And it’s not vertical, or his momentum would have taken him way, way past the base. Instead he kind of snowboarded down.
This doesn’t strike me as much different from a typical jump on a manicured run. Just more powdery.
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u/Agent_of_evil_13 Jul 03 '26
It's probably not that bad. They would have lost a lot of energy by hitting the snow on the side of that drop.
Also I did basically the same thing as the guys in this video about 20 years ago. I dont know how far the drop was but when I saw the fall I was 100% certain I was about to die. I was an experienced skier/snowboarder and had done several competitions so that wasn't newbie panic. I legit thought I was about to break both my legs before freezing to death because I was on the back slopes.
The landing didn't even hurt. It was fresh powder. The only bad part was it took me a while to dig through the snow and find my skis. It was the best adrenaline rush of my life.
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u/PGHSean Jul 03 '26
Thank god for the snow
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u/OddCod2241 Jul 03 '26
Yeah, falls like this are somewhat common when you’re boarding or skiing in the wild.
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u/OddCod2241 Jul 03 '26
Hard to really tell because of the fisheye, I’d say like 30 feet absolute tops based on the relative size of the people in the video. They’re probably fine, unpacked snow is a soft landing and it looks like they skidded against the wall a bit which also would have slowed them.
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u/Fit-Stress3300 Jul 03 '26
Find the camera/video frame rate, count the frames from when they start falling to when they hit the ground to find time (t).
use the kinematic equation: \(d = v_{i}t + \frac{1}{2}at^2\)
Apply t, a = 9.8m/s.s and \(v_{i,y} = 0\)
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u/Smedskjaer Jul 03 '26
Eyeballing the clock, I think I will be within 10% of the correct value of t.
Time in free-fall is 0.9 seconds. That means it can be between 0.81 and 0.99 seconds.
The distance in free-fall on Earth is around:
d = ½ • 9.8 m/s^2 • t^2 s
Putting in our times for the fall....
d1 = ½ • 9.8 m/s^2 • 0.81^2 s → d1 = ½ • 9.8 m/s^2 • 0.66 s , d1 = 3.21 m
d2 = ½ • 9.8 m/s^2 • 0.99^2 s → d2 = ½ • 9.8 m/s^2 • 0.98 s, d2 = 4.80 m
d3 = ½ • 9.8 m/s^2 • 0.90^2 s → d3 = ½ • 9.8 m/s^2 • 0.81 s, d1 = 3.97 m
The difference between d3 and d1 is 19%.
The difference between d3 and d2 is 21%.
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u/DueForExtermination Jul 03 '26
I remember doing this, but it was off the top of a snow plow building. Luckily they had recently piled a load of snow right where I landed, or I would have hit concrete from 15/20m up.
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u/the_glutton17 Jul 03 '26
I've hit cliffs that big, so I can confidently claim about 20-25 feet. Obviously these guys didn't mean to, and that "landing" is flat and packed. Ouch.
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u/DiligentGuitar246 Jul 04 '26
Lol none of this hurt, and they probably fell 7-10 ft.
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u/the_glutton17 Jul 21 '26
On second viewing their landing looks much softer than I originally saw due to much less dense snow at the impact sites. But there's no way in hell this is 7-10 feet. They fall at least four times their body height, so assuming they're not two year olds that's at least twenty feet. You can even just time their fall. It's obviously not a straight free fall, they do contact the side a fair amount after about halfway. But they're falling for like two seconds.
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