REALLY OT Physics formula needed | The Boneyard

REALLY OT Physics formula needed

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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.
 

eebmg

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Well, do not have paper and pencil but in steps by decomposing vertical and horizontal

1) Determine time that gravity takes in making the fall from height H (from 0 initial velocity) H=100=1/2*g*t^2 so that gives t by simple algebra. g is the gravitational acceleration (9.8 m/s^2)

2) Now the horizontal displacement in that time t is x=24=v0x*t since no acceleration due to gravity occurs in horizontal direction

Oops. Be careful. Just noticed you are using feet whereas g is in meters so you should convert all feet to meters to be consistent with the rest of the scientific world
 

temery

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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.


If I remember my college physics, it goes something like, "delta V, something something ... you should switch majors before you are in over your head. Oh, BTW, you are in over your head."

30 years of teaching biology and chemistry, and I still think it's the best advice I ever received from a teacher.


 

eebmg

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If I remember my college physics, it goes something like, "delta V, something something ... you should switch majors before you are in over your head. Oh, BTW, you are in over your head."

30 years of teaching biology and chemistry, and I still think it's the best advice I ever received from a teacher.



Some of us like the order and reduction to simple rules in Physics and do not care for the heavy memorization inherent in biology or chemistry. Thank god the world has different strokes for different folks
 

Bama fan

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Or ..... find a dozen or so idiots and use the trial and error method. :confused::eek:

You'll need a speed gun, a tape measure and a couple of paramedics.
You will also need a few undertakers. I mean anything for science, but this is a bit much! :eek:
 

Bama fan

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Some of us like the order and reduction to simple rules in Physics and do not care for the heavy memorization inherent in biology or chemistry. Thank god the world has different strokes for different folks
And thank god also for these guys:

 
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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.
Try this website.
seniorphysics.com
And search for this article.
Forensic Physics 101:Falls from a height
 
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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.

Assuming you jump out from the roof, i.e. no angle in the initial jump, I think the formula for the initial velocity is going to be roughly v ~= square_root ( (32 * x*x) / (2*y)) where x is the distance from the base and y is the distance from the top all in ft, 32 is ~ gravity in ft/sec^2. So in your case, it is going to be something v = sqrt((32*24*24)/(2*100)) ~= 9.6 ft/s, I think.
 

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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.
First and foremost, one must follow the Boneyard conventions and convert the vertical and horizontal measurements of length/height into the standard Hartley Units. All Boneyard calculations are done using those specific units, and this habit must be strictly adhered to. If we are asked to do insane things, we must have clearly defined rules! ;)
 
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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.
You can stay in feet, if you know that gravity is 32.2 ft/s/s; time is simply the SQRT of twice the height/'g' ==> (200/32.2)^(0.5), then divide that time into 24; 24/2.492224 = 9.62995 ft/s; them multiplying by 3600 s/h, and dividing by 5280 ft/mi. gives you 6.565877 mph which is a gentle jog. If you are looking for proper sig figs, s = 6.6 mph. EZPZ, lemon squeezy....
I am sure you could muster it. I am more concerned about arresting your downward motion....
Gravity always wins, even if you don't believe that it is a thing.....
 
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First and foremost, one must follow the Boneyard conventions and convert the vertical and horizontal measurements of length/height into the standard Hartley Units. All Boneyard calculations are done using those specific units, and this habit must be strictly adhered to. If we are asked to do insane things, we must have clearly defined rules! ;)
I'm sorry, that would be 1.699402941 Hartley Units/second [new offical abbreviation: HU/s) and proper Sig Figs would result in 1.7 HU/s, now isn't that relaxing...... : )
 
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Thanks for getting me warmed up! Kinematics starts for me and my dual credit calculus-based Physics students in 10 days! We don't memorize equations, we just call them by descriptive names and practice enough to engrain the relationships upon their long term memory. The two relationships are called "Truth" and the "Dirt" equation. "Dirt" is Distance = Rate X Time; distance or displacement don't care for the first two days to let it settle in with a lab. Vocabulary must be applied vis a physics framework, BEFORE applying math to solve.
One of my students was the 2019 World Champion in Physics (Rose-Hulman, Terre Haute, IN was the location, 95 miles away, kind of home field advantage) BEFORE he entered the University. Kind of a proof of concept for my way of teaching Physics in High School: Honors Physics as a Junior, ACP Physics P221 as a Senior, concurrently with AP Calculus B/C. It helps that our senior Math Teacher's children have graduated form Rose-Hulman, so we are used to teaching at a high level of rigor. Rigor, NOT rigor mortis - it has to stay fresh and engaging. Stay brilliant, my friends!
 
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Not an answer to the question, but an interesting physics anecdote.

Years ago I was designing an indoor softball field in a Hartford suburb which had a strict height restriction. I appeared before the Planning and Zoning Commission to demonstrate our need for a height increase variance. I found and cited a college project made by two Trinity College students who developed a formula and a model to determine the average and maximum height of a batted softball (55 feet).

The head of the commission, before he voted, asked me if I knew what grade was given to the students.
 
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Bama fan

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I'm sorry, that would be 1.699402941 Hartley Units/second [new offical abbreviation: HU/s) and proper Sig Figs would result in 1.7 HU/s, now isn't that relaxing...... : )
Nice work. The official UConn Bureau of Weights and Measures has accepted your conversion for consideration, An official review will be conducted upon resumption of the academic year. It is far too soon to be "relaxing", but your research holds promise. Too bad Molly graduated already, as she was the team mathematician. :) Oh , and Go Huskies!
 
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Thanks all for the responses. I have to tell you up front that my math skills are somewhere between nil and not very much. I watched the Kahn Acadamy vid that @temery linked and it directly addressed the problem I asked, as did @eebmg and @CDR-ret. Unfortunately my brain has a hard time doing anything with the information.

The question arose from a Netflix Unsolved Mysteries episode I watched last night. The scenario is this: a young athletic guy, 6'5" and 250, gets a phone call, runs out of the house and disappears. 8 days later his body is found in a conference room of a hotel, badly decomposed. He arrived in the conference room via falling through a metal roof and ceiling, apparently feet first as his legs were badly mangled. The hole he made in the roof was small and he must have been nearly vertical when he impacted the roof.

The hotel had been converted to condos, and is 13 stories tall - say 130 feet. The conference room building was located inside of a U shape of the hotel/condo building and was only 1 story. The hotel building has a large flat roof, with no retaining wall, just a sheer drop. At the 11th floor - say 111 feet up - is a narrow ledge that overlooked the conference structure. from all 3 sides of the building, accessible from condo units. The hole in the conference room roof was 45 feet from the main part of the hotel and probably only 20 feet from the wings.

Apparently no one who lived or worked in the hotel saw the deceased or heard his hitting the conference room roof. It was possible to get on the hotel roof, although not easily. There was a door to the roof (which you had to know the route to) that was generally locked, but not always. So the question is, did this guy find his way to the roof and launch himself, either from the main portion of the building (45 feet to impact) or a wing (20 feet to impact), or get onto the ledge from someone's condo in a wing and jump. And if so was it possible to go the distance needed to get to the impact site. From what I gather, it was more than possible from either location.
 
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h=16t^2 so in your example you would be airborne for EXACTLY 2.5 seconds. Horizontally, d = vt so you would need to run 9.6 ft/sec to land 24 ft out.
 
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Any discussion of Hartley Units makes me think of smoots - as in Olivor Smoots, a member of a MIT fraternity many years ago. After an evening of much liquid refreshment his brothers laid him end to end across the Harvard Bridge. Smoot was 5' 7", so that length is the official smoot. The fraternity brothers submitted a report to the Boston Police saying that the bridge is 364.4 smoots long. To this day the bridge is marked off in smoots, and if you get in an accident on the bridge the accident report states the location in smoots. Refer to the Wikipedia article for details and photos.
 

Bama fan

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Hey, is that how Ken Burns got his start in entertainment - as part of Sly's family?
Jerry Martini bears a good resemblance to Mr Burns in that photo. Great player who performed with nearly every musician you can think of.
 

Plebe

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Would any of you know the formula that would let me compute how fast I would have to be going if I jumped from a given height to land a given distance away? That is, if I was on a roof 100 feet high, and by running straight off the edge wanted to land on a spot 24 feet from the base, how fast would I have to be going when I left the roof. Disclaimer: I'm not going to do this and neither should you.
I found this page that poses the question in terms of firing a gun from the top of a building, where velocity is known but height and distance are unknown, but I believe the same principles would apply. Dr. Donovan's explanation is rather over my head, but he seems to know what he's talking about.

Disclaimer: I'm not going to do this either, and neither should you :)

 

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As I went through the sciences (my relatively unused degree is in Chemistry) I discovered that each branch (including mathematics) produced unique teachers.

In Physics, in high school, there was a very popular teacher known for, I kid you not, Harry Harry jokes. In case you don't know Harry Harry had two heads. What this had to do with Physics???

In college, one of my Physics professors was a German of the WWII generation, brilliant in atomic energy, apparently, but taxed with explaining the right hand screw thread convention to a class. Which involved making a student stand in front of the class, point his arms, and rotate. With accented commentary. Hence, 45 years later, it is still memorable.
 
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Thanks all for the responses. I have to tell you up front that my math skills are somewhere between nil and not very much. I watched the Kahn Acadamy vid that @temery linked and it directly addressed the problem I asked, as did @eebmg and @CDR-ret. Unfortunately my brain has a hard time doing anything with the information.

The question arose from a Netflix Unsolved Mysteries episode I watched last night. The scenario is this: a young athletic guy, 6'5" and 250, gets a phone call, runs out of the house and disappears. 8 days later his body is found in a conference room of a hotel, badly decomposed. He arrived in the conference room via falling through a metal roof and ceiling, apparently feet first as his legs were badly mangled. The hole he made in the roof was small and he must have been nearly vertical when he impacted the roof.

The hotel had been converted to condos, and is 13 stories tall - say 130 feet. The conference room building was located inside of a U shape of the hotel/condo building and was only 1 story. The hotel building has a large flat roof, with no retaining wall, just a sheer drop. At the 11th floor - say 111 feet up - is a narrow ledge that overlooked the conference structure. from all 3 sides of the building, accessible from condo units. The hole in the conference room roof was 45 feet from the main part of the hotel and probably only 20 feet from the wings.

Apparently no one who lived or worked in the hotel saw the deceased or heard his hitting the conference room roof. It was possible to get on the hotel roof, although not easily. There was a door to the roof (which you had to know the route to) that was generally locked, but not always. So the question is, did this guy find his way to the roof and launch himself, either from the main portion of the building (45 feet to impact) or a wing (20 feet to impact), or get onto the ledge from someone's condo in a wing and jump. And if so was it possible to go the distance needed to get to the impact site. From what I gather, it was more than possible from either location.
Absolutely possible, and probable unless his legs were injured prior to going through the roof.
 

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