A body is in freefall. Assuming normal gravity, how much distance will the body drop after, say, twenty seconds? Thirty seconds? I know the basic formula is 32 feet per second/per second, but math was never my strong suit, plus there’s probably other aspects of physics I’m overlooking.
Or, put another way, a body dropped from about a thousand feet high: How long until it hits the ground?
PAD
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I don’t have the ability to look it up myself at the moment, so I’ll give a quick remimder to whatever math genius picks this up. Remember that, on Earth, a falling man will reach terminal velocity pretty quickly. I’ve got RPG sources that say terminal velocity will be reached by a falling man after about 5 seconds, but those aren’t very stellar sources.
I now turn you over to the physics geek that will certainly answer this before I can look it up tommorrow.
If we set the starting height at x=0 and the initial velocity = o, then the formula is x=-1/2gt^2 where g = 32 ft/s^2. So, at t=20 sec, x = 6400 feet downward. At t=30 sec, x = 14,400 feet.
Solving for t, we get t = sqrt(-2x/g). So, if the top is x=0 and the bottom is x=-1000, we get t= 7.9. So, it would take about 8 seconds to hit the bottom after being dropped from a height of 1000 feet.
That’s my back of the envelop calculation. Hopefully, someone can double check my calculations.
Peter, is this for a story, or are you just bored right now? 🙂
Just did some research on terminal velocity.
Den, those figures are correct in a vacuum with no atmosphere to create drag. A parachutist with his arms & legs spread will top out his falling velocity at 55 meters/second (c125 mph). Such a person would take about 8.34 seconds to fall 1000 feet with an initial velocity of 0.
I didn’t find figures for someone flailing about in a panic, but I going with the unscientific assumption of a max velocity of 60 m/s. This takes the time down by only .02 seconds.
I made an simplifying assumption that the victim of this little though experiment would accelerate at a full 9.8 m/s/s (32 ft/s/s) until they hit terminal velocity. The fact that their acceleration would decrease as they approach terminal velocity is more than I can remember from a single college physics course, twenty years ago.
However, net answer is a bit over eight seconds to fall 1000 feet.
I did the slow subtraction from 1000 (-32, -64, -96, etc): About 8 seconds, which according to wikipedia, will reach 90% of terminal velocity.
Den seem to be on the money – though here is a less complicated way of looking at it.
The falling body is movig at 32 feet per second in the 1st second, 64 in 2nd, 96 in 3rd, 128 in 4th, 160 in 5th, 192 in 6th, 224 in 7th & 256 in 8th.
32 + 64 + 96 + 128 + 160 + 192 + 224 + 256 = 1152 feet at the 8th second. so it is slightly less than 8 seconds.
It also depends on how fast one is falling. Someone wearing a parachute will fall slower than someone who isn’t. Then there is the question on how long it takes to reach terminal velocity.
Den, your formula confuses me. Could you put in a key so that we know what everything is? Height was labeled x, but velocity wasn’t labeled, and I’m having trouble with what g is. T is obviously time, so that’s ok. Is g supposed to be velocity?
Thanks
David, you are correct that my equations ignore atmospheric drag. Of course, Peter didn’t give us any details about the nature of falling body, whether it was even a human being or just a bowling ball, so I just took the simplest assumptions and ignored atmospheric drag.
Since your example of someone spreading their arms and legs out added only about 1/3 of a second to the drop time, it’s not really that huge of a simplification.
I’m assuming that Peter wants this for a story, so “about 8 seconds” is probably a good ballpark number of his purposes. So, unless he’s actually got some resert project this weekend* where he needs a more precise drop time down to a fraction of a second, that’s what we call good enough for government work.
*Then again, maybe he is working on something. Did you meet up with the Mythbusters gusy at Dragon Con, PAD? If you did, tell them I loved the episode where they tried to slow Buster’s fall using a piece of plywood.
x=height
t=time
g=acceleration of gravity
v=velocity*
*Since I assumed an initial velocity of zero, velocity drops out of the equation: x = -(1/2)gt^2 + v0t + x0
Where v0 is the initial velocity and x0 is the starting point.
BTw, the formula for velocity is just the derivative of the above equation:
v = -gt + v0.
Assuming again, no initial velocity, the velocity when our theoretical body goes splat would be approximately 256 ft/s. This again neglects air resistance.
Thanks, this makes much more sense.
Now to throw another variable into the mix. I wonder if weight would be a factor. If I remember correctly, it shouldn’t since all objects will fall at the same rate. My question I guess would be, will they make contact with the ground at the same time?(Cats not counting since they seem to defy the laws of physics most of the time)
Sandahl, you answered your own question. Aristotle was cited saying that the heavier object would fall faster, but Galileo’s demonstration was cited establishing the two objects would land at the same time.
Sorry for rambling then.
Den,
I agree that “around eight seconds” is probably good enough for whatever Mr. David is working on, but having put the work in, I decided that I’d post my answers, anyway. Didn’t want to feel like it had all gone to waste.
Eight seconds to fall from 1000 feet.
Makes them cowboys who take that long to fall from a bull seem like wusses. 🙂
I’ve always wondered about that though – objects falling at the same rate. Wouldn’t a heavier object have to fall faster? If acceleration [falling] is proportional to force, and gravitational force is a function of the two masses and the distances between them, then a heavier object should appear to fall faster [technically, the extra acceleration would be from the Earth’s movement, but that would be a matter of perspective]. Of course, this would only be noticeable if you were dropping something as massive as the moon. An object, around half of Earth’s mass should fall [assuming about 6400km between their centers of mass] around 12 or 13 m/s/s – If I did my calculations right.
So the arguments with Byrne have lead to this, have they?
🙂
In a vacuum, objects fall at the same rate. It’s wind resistance that’ll slow the velocity. Which is why there’s that video of a feather and a bowling ball landing at the same time when dropped inside a vacuum.
Laden or unladen?
African or European?
I think other people have covered the “how long to fall 1000 feet?” question (assuming that the person starts out “at rest”), so I’ll just elaborate on a couple of other points.
sandahl – to give a more general version of den’s formula, the relevant equation is “x = ut + 1/2 at^2”.
x = distance travelled (in metres).
u = initial velocity (in metres/second).
t = time taken (in seconds).
a = acceleration (in metres/second/second).
You can use this to work out things like “if a car goes from 0-60 in 2 seconds, how far has it travelled in that time?” For falling, you just say that a=g, i.e. acceleration = gravity = 9.8m/s/s.
Scavenger – it’s funny you should mention Byrne, since I picked up his FF Visionaries v6 TPB yesterday. This included a scene where the Beyonder jumps out of a window and Luke Cage jumps out after him. Quoth Luke: “I’m heavier’n him, so I should fall faster! Once I catch him maybe I can break his fall somehow!” Mind you, it didn’t work, so this may have been a deliberate error for the character rather than the writer, since Cage also got confused about the gold standard, and had to be corrected by Iron Fist. (Also in fairness, that was from Secret Wars II #2, written by Shooter rather than Byrne himself.)
Coming back to PAD’s original question (distance travelled in 20 or 30 seconds), if we again assume that the body starts at rest, and also ignore terminal velocity, then:
When t=20, x = 1960m = 6430ft = 1.2 miles.
When t=30, x = 4410m = 14468ft = 2.7 miles.
Terminal velocity is a bit beyond what I covered in my maths/physics education. I’ll assume that the person has been tied up (and is unconscious), so we can take the maximum speed from the Wikipedia page that Mike linked to above, i.e. 320km/hour = 88.9 m/s (let’s round down to 88.8 for the “Back to the Future” reference). This raises the question of “how long will it take to reach terminal velocity?” I.e. will the person actually be accelerating for 20/30 seconds, or would they be accelerating for 5 seconds and then travelling at a constant velocity after that?
88.8/9.8 = 9.06 seconds. (This confirms that terminal velocity can be ignored for the 8 second fall that people discussed above.) For simplicity, let’s call this 9 seconds. In those 9 seconds, x = 0.5 * 9.8 * 9^2 = 396.9m.
When t=20, that gives 11 seconds of travel at 88.8m/s. This comes back to basic maths (“speed = distance/time”), i.e. x = 88.8*11 = 976.8m.
Similarly, when t=30, we have 21 seconds of travel at a constant velocity, so x = 88.8*21 = 1864.8m.
Combining the two parts:
After 20 seconds, the body would drop 1374m = 4507ft = 0.9 miles.
After 30 seconds, the body would drop 2262m = 7420ft = 1.4 miles.
Based on the phrase “the mile high club”, I guess that’s a plausible height for planes to be flying at.
The basic formula for distance, starting from rest, is d = 16 * t * t, where t is in seconds and d is in feet.
However, do to drag from air (I’m assuming he’s falling in normal earth atmosphere), a human skydiver reaches a terminal velocity of 176 ft/s in 5.5 seconds. Thus, the formula would be
d = 484 + 176 * (t – 5.5)
So in 20 seconds, a body would fall 3036 feet. In 20 secods, a body would fall 4796 feet.
A body would fall 1000 feet in about 8.5 seconds.
The basic formula for distance, starting from rest, is d = 16 * t * t, where t is in seconds and d is in feet.
However, do to drag from air (I’m assuming he’s falling in normal earth atmosphere), a human skydiver reaches a terminal velocity of 176 ft/s in 5.5 seconds. Thus, the formula would be
d = 484 + 176 * (t – 5.5)
So in 20 seconds, a body would fall 3036 feet. In 20 secods, a body would fall 4796 feet.
A body would fall 1000 feet in about 8.5 seconds.
I couldn’t get to the site for a while (anybody else seeing this problem, too?), but now I figured I’d weigh in on a few things. The numbers (roughly 8 seconds, with or without air resistance) all look fine.
First… one comment.
Den seem to be on the money – though here is a less complicated way of looking at it.
The falling body is movig at 32 feet per second in the 1st second, 64 in 2nd, 96 in 3rd, 128 in 4th, 160 in 5th, 192 in 6th, 224 in 7th & 256 in 8th.
That’s not entirely correct. The falling body is moving at 32 feet per second AT THE END OF the first second. During the first second, it’s sped up from zero to 32 ft/s. Similarly, during the 2nd second, it speeds up from 32 to 64 ft/s. Therefore, if you want to do the adding up of distances you were trying you’d have to use average values: 16 feet in the first second (average of 0 and 32), 48 in the second, etc. You get an answer that’s close to yours, but not quite.
I’ve always wondered about that though – objects falling at the same rate. Wouldn’t a heavier object have to fall faster? If acceleration [falling] is proportional to force
It’s proportional to force, yes, but it’s also inversely proportional to the mass of the object being pulled. (Consider throwing a lasso around and pulling on a roller-skating gerbil vs. a truck.) Since the object’s mass is both directly AND inversely proportional to the acceleration, then, it cancels out and the acceleration is mass-independent.
(If you bring in air resistance, the force of which doesn’t depend on mass, then mass becomes a factor.)
Now I just want to know why Peter needs to know this! 🙂
TWL
I seemed to tremember that a good ballpark estimate for terminal velocity is 120 MPH in a prone position and 180 MPH in an upright position, because the same mass with less wind resistance will fall faster. But, for questionsa like this, why not check wikipedia?
http://en.wikipedia.org/wiki/Free_fall
This provides the equations for free fall, and includes a link to
http://bpesoft.com/s/wleizero/xhac/
which can be used to plug in your own numbers in a real-world situation including air resistance, etc. Using the skydiver free fall, from 1000 feet to sea level gives a fall time of 9.271 seconds and an impact velocity of 50.18 m/s (152.95 MPH).
Peter, this thread has provided you with more information than you ever wanted to know about free fall, hasn’t it?
I don’t need math. I just jumped out of an airplane at 1000 feet w/ a stop watch. It took me about 7.5 seconds to hit the ground, because I’m fast. I landed in some tall grass, so I’m fine. I hope this helps.
Mass does factor into it because it isn’t falling through a vacuum – assuming this takes places on Earth – and if a hot air updraft can keep a glider aloft for hours, it ought to measurably affect a falling body, depending on its mass. Enough to make a difference in the story? That’s another matter.
QUICK! SOMEBODY CALL A PHYSICS INSTRUCTOR!
It’s been long decades since I had physics class, but I seem to remember that, because of resistance, a falling human body tops out somewhere around 100 feet per second. But I can’t remember for sure.
I also seem to remember 16′ 1st second, 32′ 2nd second, 64′ 3rd second, around 100′ 4th second, 100′ the 5th second, etc. That would seem to me to add up to around 12 seconds.
Falling 1000′ while conscious would probably subject the person to time distortion, meaning it would seem all too short, yet an eternity.
Being nekkid might speed the fall slightly because of less drag.
For extra credit, who would fall faster? Calista Flockheart or Dolly Parton?
🙂
The formula for terminal velocity involves density (about 1 for most people, but less if you have baggy clothing) and cross-sectional area (lower if you spread-eagle, higher if you arrow down) and the viscosity of the air, among other things. 120mph or 60m/s are cited as generic “most of the time it’s around this” levels, since most people don’t want ot mess about with differential equations.
Suffice to say, after about 200-220 meters, someone who isn’t TRYING to fall really fast will hit terminal velocity. Someone who IS trying to fall fast, by diving, probably won’t hit terminal velocity before hitting the ground if they start 300 meters (1000 feet) up. Numbers for each case already exist in this thread. 🙂
(For some reason, I got a blank page for here all yesterday, or I would have responded sooner.)
After reading David Van Domelen’s post, I wonder why the rest of the world hasn’t switched over to feet instead of that hard-to-covert-to-feet meters crap.
😉
QUICK! SOMEBODY CALL A PHYSICS INSTRUCTOR!
You’ve got at least two in this thread already — me, and David Van Domelen. (Den teaches in the sciences as well, though not physics specifically if I remember right.) How many more do you want? 🙂
And while I agree with Dave’s “most of the time it’s around this” numbers, I tend to give a couple of others to my classes when this comes up. If you’re falling spread out like a flying squirrel and have lots of baggy clothing, then it’s probably more like 70-80 mph (less if you have a parachute, obviously). If you’re wearing Spandex and deliberately falling like a dart, then it’s significantly higher than 120 mph — it’ll depend on your weight, but it’s a lot closer to 200 than it is to 120. Impact at the first speed (70-80 mph) is survivable with good ground conditions and significant good luck. Impact at the second speed typically requires that friends arrive with mops and squeegees. 🙂
TWL
Okay, all you physics guys out there–would the horizontal velocity change the vertical velocity any? IE, jumping from a speeding plane rather than from a balloon or a cliff? I’ve halfway convinced myself that all it would do would be to change the location of contact with the surface, but there’s something tickling me in the back of my head, and I just really hope it isn’t a flea.
Tim, when you give that problem to your class, do you add that being spread out like a flying squirrel may lead to later problems with two Russian agents and a silly moose?
How baggy do your clothes have to be to add a significant difference in the time of the drop? (I swear the first one to mention parachute pants gets sixty lashes with a wet noodle.) Wouldn’t moving through the air just push baggy clothes up against you, conforming to your shape, and making them come to a point over you? In my head, I’ve compared this to swimming in baggy clothes versus a wetsuit, since contrary to my wife’s wishes, we’ve never skydived but I’ve been snorkeling since I was 11. Is this a valid comparison or am I completely off on this?
Hmmm. Stace wants me to jump out of a plane. Should I worry?
I didn’t find figures for someone flailing about in a panic, but I going with the unscientific assumption of a max velocity of 60 m/s. This takes the time down by only .02 seconds.
When I did my one and only jump the instructor told me that a controlled power dive – arms tucked in tight to the body and legs together – would result in falling at a terminal velocity of about 200mph. The flailing panic fall will certainly be slower.
I have a question, is it possible to break terminal velocity, if a person adds some type of thrust to their fall, like Superman flying after Lois after she’s been thrown out a window, again.
Posted by Sean Scullion at September 24, 2006 Posted by Sean Scullion
Okay, all you physics guys out there–would the horizontal velocity change the vertical velocity any? IE, jumping from a speeding plane rather than from a balloon or a cliff? I’ve halfway convinced myself that all it would do would be to change the location of contact with the surface, but there’s something tickling me in the back of my head, and I just really hope it isn’t a flea.
Not in general, unless it’s a Very High horizontal velocity that carries you significantly over the curve of the Earth -up to the point where you hit orbital valocity and really are doing what Douglas Adams describes as “the way to fly” – hurling yourself at the ground and missing.
As to terminal velocity – are we calculating the time to reach V-sub-t simply as “so many seconds at 32 ft/sec/sec”, or allowing for the reduced acceleration due to air drag?
Oh, since someone mentioned the gold standard upthread, and we’re discussing if heavier masses fall faster or not – whichg weighs more – a pound of gold or a pound of feathers?
Horizontal velocity has an affect on vertical velocity only in so much as the Earth is curved and a truly straight line horizontal movement eventually leads AWAY from the Earth’s surface due to its curvature. If the “horizontal” flight was a CURVED flight arc that mirrored Earth then it would be an orbit.
As for the base question
t = time (in seconds), g = -32 ft/sec/sec, then where
v = velocity, v0 = initial velocity (when t = 0)
v = gt + v0
h = gt(t+1)/2 + v0t + h0, thus:
The table really simplies this if you don’t want to do the higher order math.
t a v h
0 -32 0 0
1 -32 -32 -32
2 -32 -64 -96
3 -32 -96 -192
4 -32 -128 -320
5 -32 -160 -480
6 -32 -192 -672
7 -32 -224 -896
8 -32 -256 -1152
9 -32 -288 -1440
Where v0 = 0 and h0 = 1000 then
h = gt(t+1)/2 + v0t + h0 = gt(t+1)/2 + 1000
= (gt^2)/2 + gt/2 + 1000
= (-32t^2)/2 + (-32)t/2 + 1000
= -16t^2 – 16t + 1000
When the body hits the ground, h = 0, so
0 = -16t^2 – 16t + 1000
16t^2 + 16t = 1000
t^2 + t = 62.5
t^2 + t + 1/4 – 1/4 = 62.5
t^2 + t/2 + t/2 + 1/4 = 62.5 + 1/4
t(t + 1/2) + (t + 1/2)(1/2) = 62.5 + 0.25
(t + 1/2)(t + 1/2) = 62.75
(t + 1/2)^2 = 125.5/2 = 62.75
t + 1/2 = square root(62.75) = sqrt(62.75)
t = sqrt(62.75) – 0.5
t ~ 8-0.5 ~ 7.5 seconds
— Ken from Chicago
Calista and Dolly would fall at the same rate, but Dolly would fall more often given the bounce factor. The real math comes into plotting the declining height of each bounce.
Falling from 1000 feet wouldn’t be enough to kill all that much horizontal velocity.
And a pound of feathers weighs more than a pound of gold. But they use different pounds and different ounces (an ounce of gold weighs more than an ounce of feathers, but there’s 16 ounces of feathers in a pound, and only 12 ounces of gold in a pound).
Okay, all you physics guys out there–would the horizontal velocity change the vertical velocity any?
As was said before I got here, nope — not unless you’re dealing with the curvature of the Earth. (Here’s a mind-blower: if you’re dealing with the curvature of the Earth, the main reason vertical V gets affected is that the definition of vertical changes as you go. I always have fun asking students that one.)
Now, technically, since air resistance is dependent on your overall velocity, I believe that you might increase the drag somewhat by leaving with a high horizontal V, but my sense is that it won’t make a significant amount of difference.
Tim, when you give that problem to your class, do you add that being spread out like a flying squirrel may lead to later problems with two Russian agents and a silly moose?
Alas, no. I’ve decided in advance that they almost certainly wouldn’t get it, and I don’t need to be known too strongly as “that weird guy in the basement.” (My classroom is in the basement, so the latter part is literal.)
I have a question, is it possible to break terminal velocity, if a person adds some type of thrust to their fall, like Superman flying after Lois after she’s been thrown out a window, again.
Absolutely. Terminal V is just defined as the velocity where the upward force of air resistance balances the downward force of your weight. You can certainly be falling faster than that, even without thrust — it’s just that without thrust, the net effect of the two forces will be to gradually slow you back to terminal V. If you’re willing to spend the energy to provide a sustained thrust, you can achieve pretty much any reasonable speed you want.
TWL
Impact at the first speed (70-80 mph) is survivable with good ground conditions and significant good luck. Impact at the second speed typically requires that friends arrive with mops and squeegees. 🙂
So I’m falling out of an airplane over water. It doesn’t matter how it happened. OK, I got too insistant on an extra packet of nuts, happy now? Anyway…is my best strategy to spread out like a flying squirrel for the majority of the trip down and then switch to the dart position at the last possible second?
I know that I’d also need to clench my buttocks as much as possible so that my large intestine doesn’t shoot out my mouth from the World’s Worst Colonic…I’m thinking my butt will be plenty clenched at this point.
You don’t really want to hit water at all…you have to move your mass in water out of the way in a REALLY short time, and that tends to be lethal even if you execute a perfect dive after maxing out your air resistance until the last second.
You want to hit snow, or trees, or better yet, snowy trees on the slope of a ravine full of snow. WWII paratroopers have survived that kind of thing…it’s all about making the “slowing down” part take as long as possible.
Yeah, Tim, I was having trouble seeing the site for the past several days too.
1 Terminal velocity for a human body is right around 120 mph. Hope that helps some.
Tim,
Anyone who would tell their parents that they’re considering naming their daughter “Mothra” has already made it to weird. You’re just hiding it better than some people. The next step is to get you out of the weird closet. If we can get you confined to a garage for a reasonable period of time, I figure that we’ll have flubber on the open market within ten years and I’ll finally get that flying car they promised me thirty years ago.
And yes, I also experience problem getting onto the sight for several days.
David,
That only counts as “made it to weird” if I was serious when I told them. If we were just doing it to get a rise out of them (and Lisa participated just as enthusiastically as I did), then that doesn’t make us weird, just enjoyably sadistic towards our parents. 🙂
TWL
This is why people think Trekkies are weird, you know 🙂
Details, Tim, Details. If it means I get a flying car, I’m perfectly okay with framing you. 😉
If it means I get a flying car, I’m perfectly okay with framing you. 😉
You do realize that the above statement is permissible in court. 🙂
TWL
Okay, this is waaaayyyy off topic but too good not to share.
My uncle fought in the Vietnam War. He didn’t want to scare my grandparents, however, so he told them he was going to be stationed in Korea.
After he got back from Vietnam, he decided to marry my aunt. He brought her to meet my grandparents for the first time, so he could announce their engagement.
Upon hearing the announcement, my grandfather looked at my aunt with dead seriousness and said, “I don’t think he can marry you unless he divorces his Korean wife.”
Like I said, this was the first time my aunt had met my grandfather! My uncle had to spend some time assuring her that my grandfather was a bit of a comedian.
Tim, I think with this story I’ve got you beat. 🙂
You know, I don’t see what’s so weird about naming your kid Mothra. I mean, it WAS a girl. It isn’t like you named her Megalon or King Caesar.
I wanted to name the CAT Biollante and got overruled so you can see how far MY opinion goes…
On the other hand, I DO get away with calling my beautiful daughter Amanda “Manda”. Booyah! And nobody’s the wiser!
That’s a good story — I hope I get the same chance to weird out whoever my daughter brings home.
Straying even further off topic … I don’t think I’ll be giving my parents quite the hard time that I did with the “Mothra” thing for a while. My mother was just diagnosed with esophageal cancer earlier today, and just told us the news a couple of hours ago, so I’m in a very weird emotional place right at the moment. Anybody ’round these parts got good stories (and by “good”, I mean ones with very cheerful endings) they can share? I could certainly use ’em…
TWL