A blog after my own heart

Now that's what I'm talking about! Remember when I used to do superhero-related Fermi Problems? Well check out this out: there's this blog called Polite Dissent and last month he asked a Fermi problem: Could Superman cure cance?
I loved the post and I'm sure you will too.

Batman's Last Fermi Problem

Diamonds are just carbon all bound together. Each carbon is bonded to 4 other carbons. If you think about that a bit you will see that means there are TWO bonds for every atom (imagine/draw a fairly large grid - count the number of nodes/intersections and count the number of connecting lines. If this ratio isn't exactly half it's because the edges of the grid might cause trouble [if you draw it smarter or bigger it will get closer to 2]). That means that there is 3eV of energy available for every atom. Let's assume Mr. Freeze's gun can access all that energy.

Now that we know how much energy we can get from each atom we need to guess how many atoms are in a diamond. To do that we need the mass of the diamonds. When jewelers talk about carats their really just talking about mass --- 1carat = 0.2g. Maybe you know what the carats of some famous diamonds are? 100 carats is a BIG diamond so let's say 30 carats or in grams: 30 carats * (0.2g / 1carat)= 6g.

How many atoms are in 6g's of diamond? Well what's the atomic weight of carbon? If you don't know, can you imagine where carbon is on the periodic scale? It's near the beginning but not first and it's nowhere near the end. So say between 1 and 100. You could then say 10g / mol. OR you could have memorized your periodic table and remembered the value is actually 12 g/mol but really what's the difference?

So now we know the number of atoms in the diamond is about 6g * (mol/12g) * (6*10^{23}atoms/mol) = (36/12)*10^{23}atoms = 3*10^{23}atoms. And since we guessed about 3eV is released from each of those which means the energy is (3eV/atom) * 3*10^{23}atoms * 1.6\times10^{-19}J/eV = 9* 1.5 *10^4J = 13.5 * 10^4J = 10^5J.

Is that alot? We said in previous questions that a person uses 3.5*10^3J every day and the energy of a subway train is 10^7J. The energy Mr. Freeze can get from a diamond is somewhere in between.

And here ends the Fermi Problems part of my Batman training - at least on the blog. But if you have any superhero Fermi problems that you think I should do on Building Batman just leave a comment and I'll give them a whirl.

Batman's May Be Chemist But I'm Not Yet

Whoa!

Don't I have any chemists reading Building Batman? No? No chemists training to become Batman? Oh, well...

I was thinking about the Final Fermi Problem and it occurs to me that I made a bad estimate:
I said that each carbon is bound to 8 others because of the crystal structure. That's not true.

Carbon is tetravalent which means it will always have the equivalent of 4 four bounds, no matter the crystal structure.

Now you and I know that when estimating that factor of two isn't a big deal but I thought I'd better tell you in case you're you're doing this last Fermi question or in case a chemist stumbles past Building Batman.

PS If you are a chemist, you might get a kick out of this periodic table

Fermi Freeze - Batman Training's Last Fermi Problem

I like that fact that every reaction releases about 1.5eV of energy so let's make our last superheroic Fermi Problem one that uses that fact and then move on with the Batman Training.

What's the energy Mr. Freeze get's from diamonds (assuming it's a chemical process and not nuclear)?Diamonds are just carbon all bound together. Each carbon is bonded to 8 other carbons. (As always I'm just putting the simplest way of thinking about this together: Diamond is a crystal structure made of two fcc lattices).

Flash Forward a Billion Years

So we are going to guess how much oxygen the Flash uses each day.

As a few reminders:
   (1)The hydrocarbons in our food get split up and carbons are oxidized into C0_2 meaning we really do burn the food we eat.

   (2)Typically,there is an energy release of about 1.5eV when one atom exchanges an electron with another. The units eV are related to joules by 1eV = 1.6 * 10^{-19}J = 2* 10^{-19}J.

   (2)Food is mainly composed of long chains of CH_2. When it is decomposed to get the energy it holds the carbon binds to two oxygen and the two hydrogens bind to a single oxygen meaning there are TWO reactions for every CH_2 which need THREE oxygens.

We already found that the Flash uses 5 * 10^9J per day. How many reactions is that? Turn joules into electron volts and divide by 1.5eV for every reaction:

5 * 10^9J * (1.5eV / 1.6 * 10^{-19}J) = 5  * (1.5/1.6) * 10^{28} = 5* 10^{28} reactions.

And we said 3 oxygens for every 2 reactions means we need 5 * 10^{28} * 3/2 = 7 *10^{28}$ oxygens per day.

That sounds like a lot but is it really? There's 6 * 10^{23} atoms in every mole and oxygen is eighth on the periodic table which means 1 mole weighs 8 grams. So then the Flash goes through 7 * 10^{28}  *  (1/6 * 10^{23}) * 8g = 8 * 10^5 g = 800 kg of oxygen each day. WOW but wait a minute...

In the last question we found that there's about 10^{18} kg of oxygen in the atmosphere. Which means it would take the Flash 10^{15} days or 3*10^{12} years or 3000 billion years.

Fermi's Big Gulp

So we already guessed how much the Flash would have to eat but what about breathing? He needs air to turn all that food into energy. Would he use up the entire atmosphere?




Ok. Some backup science info. Energy release is controlled by mitochondria and the production of ATP through the Krebs cycle. It seems fairly complicated to me but in essence the most important thing for accessing all the stored chemical energy in the food is the oxygen intake and the carbon dioxide output. The hydrocarbons in our food get split up and carbons are oxidized into C0_2 and a bunch of water reactions are also needed to make this happen BUT the point is we really do basically burn the hydrocarbons that we eat.

Chemical reactions are all about exchanging electrons. Remembering that we are estimating: typically, when one electron is exchanged between two atoms there is an energy release of about 1.5eV. The units eV stand for electron volts and their just a convenient unit for \textbf{energy} when electrons are involved. To convert to everyday joules just know 1eV = 1.6 * 10^{-19}J = 2 * 10^{-19}J.

And one last thing: Most everyday hydrocarbons are just long chains of CH_2.

Breathing Room

Last night I attended my very first Jiu Jitsu class and so have raised my Batman training to the next level. I'll tell you all about it tomorrow. But tonight, let's answer yesterday's non- Superhero Fermi Problem (we'll use this answer for the next, and second from last Fermi problem).

The atmosphere pushes down on you (you might not have the weight of the world on your shoulders but you definately have the heavens weighing down on you). That pressure is measured in atmospheres (atm). The pressure of the atmosphere is normally 1atm (duh). In useful units 1 atm = 10^5 N/m^2. Where's that pressure come from? All the air molecules over head weigh down on you. Since the pressure is force per area, above every 1 meter squared there is a force (or weight) of 10^5N and remembering Force=Weight=mg where we round gravity to 10m/s^2. So the mass of the particles above 1 meter squared is 10^4kg.

We already used the area of the earth in the buses Fermi question and we use it again here: surface area of earth =5*10^{14} m^2. We know the mass of atmosphere above every square meter AND we know the total number of square meters that make up the surface of the planet. Therefore, we estimate the total mass of the atmoshere to be 10^4 kg/m^2 *   5*10^{14} m^2 = 5 * 10^{18}kg of air in the atmosphere!

BUT

Air isn't pure oxygen, O_2. The percent of the atmosphere that is O_2 definitely isn't 100%and it's not 1%. So we could guess 10% but I think that it's fairly common knowledge that air is 20% O_2 so let's use that number.

5 * 10^{18}kg * 20/100 = 10^{18}kg of oxygen. That's alot of elephants - or subway trains for that matter of oxygen. Wikipedia says that the atmosphere is really 5 quadrillion tons. A ton is 1000 kg and a quadrillion is 10^15 so we were only off by a factor of 5. Once again, we're pretty darn close.

Fermi Question: What is the mass of the Atmosphere

What's the total mass of all the oxygen in the atmosphere?

Useful unit conversion: 1atm = 10^5  N/m^2.

Would the Real Superman Please Standup

The mass of the train should be the same (recap: car of a train is heavier than a truck [we've heard the term oneton-truck so let's say a truck weighs 1 ton]. It's probably more than 10 trucks but less than 100 so we say 30 tons. And there are 6 cars so about 2 * 10^5kg.

In the Spider-man question we said the train was going about 30km/hr. But this is Metropolis, the city of tomorrow! So let's say the city has levitating Shinkansen which can travel upto 600km/hr. Since it probably didn't have time to reach it's fastest speed in the city we'll say the train has a velocity of 100km/hr = 100km/hr *  1000m/km * 1hr/3600s = 100/3.6 m/s = 30m/s when it comes flying of the tracks.

Just like before we can find it's original kinetic energy (KE = 0.5 * m * v^2 = 0.5 * 2 * 10^5kg * (30m/s)^2 =10^8J). That's an order of magnitude larger than the train Spidey stopped.

But here's the second kicker: whereas Spider-man took 10 blocks to stop the train Supes has only half a block if this train isn't going to crash into a crowded skyscraper. So remember there's 10 blocks in every kilometer - so half a block is about 50m.

So the force is F=W/d = 10^8J  / 50m = 5 * 10^6N.

That implies Superman is about 500 times stronger than Spider-man. If that's not a 'super' Fermi problem, I don't know what is.

All this talk about Superman is a good excuse for me to put up this link to a Comedy.com post

Metropolis' Marvel - Stronger than a Locomotive

This superhero Fermi problem is just like the last Spider-man one so we'll use it as warm up.

In Matt Wagner's Trinity, Superman catches an out-of-control train (with say 6 cars again) that comes off it's tracks as it rounds a corner. He brings it to a stop before it can hit any buildings.

How much force did the Man of Steel use to do this?

The Dark Knight Returns

I registered and payed for my beginner-level Jiu-Jitsu yesterday. It feels like I've been wait forever to start. The first class is on Sept. 2nd. Classes occur once a week and go for 10 weeks. I'll post what I learn each week.
P.S. I'll post all the other Fermi problems this weekend and then we'll be done with them.

Sticky Situation

This one's stolen from Guesstimation.

In the second Spider-man movie Spidey stops a six car subway train by attaching his webs to it and pulling it to a stop in 10 or 20 blocks.

How much force does he need to do this?

Food for Thought

Let's answer this in three parts. How much energy does the average American consume, how much does that cost, how much does each calorie cost and how much energy would the Flash need.

There's a number of ways we could estimate how much one person spends per month for food. You could say that you eat out every meal (3 times a day) at some fast food joint for $10 a meal. That kind of life style has a monthly bill of 30 * 3 * 10=$1000 per month. I think that's high at $30 per day. What about $1 per day? At that price food would only cost you $365 each year. Ya right. So the easy numbers between $1 and $30 are about $10 per day or $3 per day. Let's be nice to Wally and say $3 per day or about $100 per month.

So we've guessed it can costs about $3 a day to eat (I know it seems low). Now we must guess how many calories you consume a day and so find the cost of every calorie.

How many calories do you eat per day? Maybe you know this maybe you don't. If not, maybe you know how many calories are in a McDonald's Value Meal (1167 cal) or in a can of coke (200 cal) or in a steak (500 cal). Therefore, you eat more than 1000 cal and probably not 10,000 cal. So then we guess the geometric mean which is 10^{7/2} = 3000 cal per day. So let's convert from calories to Joules (there's 4200 Joules in every calorie). 3000 cal/day * 4000 J/cal = 12 * 10^6 J/day. A physicist like Ray Plamer or Reed Richards would be happier if we turned this to Watts. Watts are Joules per second. So as a side note the power usage of a human being is 12 * 10^6 J/day * day/24hr * hr/60min * min/60sec = 150 W. Oh good - more power than a lightbulb.

So anyway that was a sidequest. The important thing is that a person consumes 3000 cal or 10^7 J everyday. Plus spends $3 a day for that food. That means that $1 will buy you 1000 cal or 4 * 10^6 J.

Next we have to estimate how much energy the Flash needs. Let's guess that most of his energy goes to running (that way if our answer is small compared to a normal person we can add the energy of a normal person to it or if our answer is big compared to a normal person we know that it is insignificant). The energy of him running is kinetic. We already used the equation for kinetic energy KE = 0.5 * m * v^2. If we allow for fictitious elimination of frictitous forces then if we can guess his speed, we will know the energy needed to get him to that speed. He needs that much energy every time he runs to that speed. Remember we are guessing for an average day, not at his fastest nor his slowest.

The speed of light (3 * 10^6m/s)?. %1 the speed of light? The speed of sound (300 m/s)? The speed of thought (only about 100m/s since it depends transport mechanisms of ions crossing membranes)?

Let's guess that he goes 10 times the speed of sound for an average run. The energy he needs to reach that speed is KE = 0.5 *m * v^2 = 0.5 * 100kg * (10 * 300m/s)^2= 0.5 * 100 * 9 * 10^6J=5 * 10^8J.

Since he needs that much energy every time he gets to those speeds we need to ask how often does he go that fast? We are talking about a cocky jock with a fairly large rogue gallery. So let's say he goes more and once a day but less than 100 times a day. The geometric mean: 10 times.

Then the energy the Flash needs for running in a day is 5 * 10^9J. At a price of $1 for every 4 * 10^6 J, the Flash must spend 5 * 10^9J / (4 * 10^6 J/$1) = $10^3 = $1000 more per day than muggles.

We said that a single person could eat for $100 per month and based on that the Flash's monthly grocery bill is $30,000. That's more than some of use make in a year! Notice that if we had said that Wally ate out for every meal his bill would be 10 times greater. A bright supervillain interested in discovering the Scarlet Speedster's secret identity might begin by looking for abnormally large grocery bills.

The Cost of Being Super

How big is the Flash's grocery bill?

For this superheroic Fermi Question you'd might like to know that every Calorie that you eat has about 4200 Joules of energy. Also guessing how fast he goes is tough. Consider average days for the Flash and think to yourself that he's much faster than the speed of sound but not any where near as fast as the speed of light. (Just an average day. That's all I'm saying.)

Skeletal

How heavy is he without his skeleton? I'm about 70 kg. Logan's pretty buff. Maybe say 100 kg.

How much of that is bone? 1%? Nah. 100%? 30%? 10%? Ya, 10% seems like the best answer. 10% of 100kg is 10kg. So that's 10 kg of bone and 90 kg of other stuff in your body.

Well, let's guess that bone and muscle and everything inside of us is all the same density. This is really not so bad of a guess because nothing inside of us is as heavy as rocks or as light as air. We pretty much have the same density as water.

The density of water is 1 kg/L = 1000 kg/m^3. So then 10kg of bone has a volume of about 10 kg / 1000 kg/m^3 = 10^{-2} m^3.

If that normal skeleton is removed and replaced with an adamantium one. If we can guess the density of adamantium then we can multiply it to the volume of a skeleton that we just found and we will know the mass which we can then ad to the 90 kg of left over non-skeletal stuff.

What's the density of adamantium? Well it's got to be heavier than water. 100 times heavier? No way. 10 times heavier seems like a good guess which says it has a density of 10 * 1000 kg/m^3 = 10^4 kg/m^3. Another way we could find this is we could look up the densities of some metals and notice that tungsten is pretty heavy at 19.25 g/cm^3 and that iron and silver and all of those are all pretty close at 7.9 or 10.5 g/cm^3 which are both 10 g/cm^3 = 10^4 kg/m^3 as far as we are concerned.

So using this density we find that the mass of an adamantium skeleton is about 10^{-2} m^3 * 10^4 kg/m^3 = 100 kg. So now with his new shiny skeleton Dark Claw or Wolverine weighs 100 kg + 90 kg = 200 kg or about twice as much. This isn't a shockingly high value. A person had an adamantium skeleton wouldn't sink in and leave foot prints in concrete or be too heavy to use the elevator.

Dark Claw's Adamantium Skeleton

In the 1990's DC and Marvel did a cross over event and mashed their characters together into "amalgams". The character they combine with Batman was Wolverine to create: Dark Claw.

At that time Wolverine had an adamantium skeleton which was one of the things that helped make him invincible (Batman often suffers from the same problem: invincibility by popularity). I can imagine some side effects to having a metal skeleton. Is weight one of them?

How heavy is Wolverine/Dark Claw with his adamantium skeleton?

Tons of Kryptonite

Let's assume that the entire planet of Krypton is made of Kryptonite. With the wide variety of kryptonite this doesn't even feel like such a stretch of the imagination.

So the planet explodes causing the mass that was the planet to shoot outwards and expand as a shell of debre. If we can figure out the fraction of that shell to hit the earth than we know the amount of kryptonite to hit earth (if we know the original mass of the planet krypton. Let's go with the 10 times greater than earth that Mohitparikh guessed instead of my 25 times (although for estimating it doesn't really matter!)).

We need to know the area of the shell. That depends on how far it's expanded. So we have to ask ourselves: How far is Krypton from Earth?

Well, how long did it take Kal-El to get to Earth? The fastest his rocket (it's usually a rocket) could possibly travel is the speed of light. You don't have to know the speed of light to know that a light-year is a unit of distance. It is the distance that light can travel in 1 year. So if Kal-El was put in the rocket (that goes the speed of light) as a new born and he arrived on earth as a yearling then Krypton is 1 light-year away. If he arrives as a 10 year old then Krypton is 10 light-years away. Here's a fact: No solar system is as close as 1 light-year away from us. Usually Kal-El is not 10 years old when the Kents find him so let's take the geometric mean and say that he was a toddler of 3 when he crashed on Earth and therefore Krypton is 3 light-years away.

Some more facts: Proxima Centauri is 4.2 light-years from us. Sirius A and B are 8.6 light-years. Vega is 25 light-years. So by guessing 3 light-years we are saying that Krypton was one of Earth's closest neighbors in a really big neighborhood.

Let's find out what 3 light-years are in useful units. Everyone rememembers the speed of light? It's 3 * 10^8 m/s. Here's the coolest trick I learnt from the Guesstimation book: You can either to all the slow multiplying of 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, 365 days in a year --- OR --- you can memorize that there are pi * 10^7 s in every year. It's a really good estimate. It's easy to remember that pi is about 3.14 and there are 3.15 * 10^7 sec / year. Cool, eh?

Anyway, how many meters to Krypton? 3 light-years times 3 * 10^8 m/s * pi * 10^7 s/year equals 27 * 10^{15} m (cuz pi=3, remember?). Which we round to 3 * 10^{16} m.

So what is the area of the shell of the exploded Krypton when it reaches earth? Well, we just found the radius of the shell (3 * 10^{16} m) and the total area of a shell is A=4 * pi * R^2. Which means the area is 4 * 3 * (3 * 10^{16})^2 = 4 * 3 * 9 * 10^{32} = 100 * 10^{32} = 10^{34} m^2.

We only need one more piece of information: How much of that hits earth?

All the Kryptonite is kind of evenly spread over the sphere so the density is d = M / A = 6* 10^{25}kg / 10^{34} m^2 = 6 * 10^{-9} kg/m^2. Now we know how many kg pre square meter and all we need is the area of the Earth that this hits. It's really the silhouette area or the cross sectional area A = pi * r^2. To do this you need to know or guess the radius of the Earth. I happen to know that it is r = 6000 km = 6 * 10^6 m.

That means that the cross section of earth is A = 3 * (6 * 10^6 m)^2 = 3 * 6 * 6 * 10^{12} = 100 * 10^{12} = 10^{14} m^2. When we multiply the area by the density we will have the mass of Kryptonite that hit the Earth. 6 * 10^{-9} kg/m^2 * 10^{14} m^2 = 6 * 10^5 kg. Holy Rao, Superman. That's 600 tons! But if you think about spread over the entire earth most of it would be dust and really we never made any guess about what fraction of the mass on Krypton was Kryptonite.

Notice, the Wikipedia page on Krypton says that Superman's home planet is 50 light-years away. If we ignore what this would mean about how fast Kal-El's rocket would travel (not to mention the Kryptonite rocks themselves) and redo the calculation then there's only 2 tons of Kryptonite on Earth. BUT even if it moved at 10% of the speed of light, it would take the Kryptonite 500 years to get here.

How much kryptonite is on Earth?

Batman's best friend Superman is a allergic to green chunks of his home planet. If Batman wants to find all of the Kryptonite on earth and destroy it or lock it away to keep Superman safe, how much should he look for?

OR

Batman doesn't like the Big-Blue-Boyscout who hasn't even figured out he can fly yet. So Batman's going to collect all the Kryptonite in the world to line the Batcave with thus making it impossible to Superman to attack Batman's child kidnapping and government over-throwing operations. How much Kryptonite is available?

This Fermi problem was posed by the book Guesstimation.

Crazy Krypton

The mass of a planet creates gravity. If two planets are the same size and one has twice the mass as the other than it will also have a gravitational acceleration that is twice as big. On earth the acceleration of gravity is g=9.8ish m/s^2. For guesstimating 9.8 is rounded to a nice g=10 m/s^2.

How high is a tall building? 1 story? Nope. 10 stories. No. 100 stories. Yes 100 stories (the Chrystler Building is 77 stories and the Empire State Building is 102 stories. How high is each story? 9ft.

So that means that the total height of a tall building is 900ft. To make ft into meters we multiply by 0.3 making a tall building 300m tall.

Relative to the ground, the potential energy at the top of the building is PE = m*g*h. All we need to do is guess the mass of Superman. Let's use a nice round 100kg (He's not 10 kg and he's definitely not 1000kg (that's a ton) so the geometric mean is 100kg). So at the top of his jump Supes' potential energy is 100kg * 10m/s^2 * 300m = 3 * 10^5 J (J=joules which we can think of as shorthand for all those more basic units which went into the calculation or we can just leave them whatever).

So then in order to get that high he must start out with a kinetic energy which is that big: KE=PE. 1 / 2 m*v^2 = 3*10^5kg m^2 / s^2. We said Superman's mass was 100kg so the velocity is v=sqrt(6*10^3). The square root of 10^3 is 30 because we divide the exponent 3 in half and multiply the coefficient by 3 for the 0.5 that is left in the exponent. The square root of 6? I don't know but I know that the square root of 4 is 2 and the square root of 9 is 3 and 6 is between 4 and 9 so I'll say 2.5.

So Superman's speed when he jumps must be v=2.5 * 30 = 75m/s. That's pretty fast but not nearly as fast as sound (340m/s) so he doesn't have to worry about causing sonic booms. How about as fast as a bullet? Not a chance Check out the Orders of Magnitude wikipedia page. A bullet moves at like 3 times the speed of sound or 1000 m/s.

So if he jumps at 75m/s, what kind of force must his legs exert to get that fast? F = m * a and he starts from standing still and gets to 75m/s. The acceleration is a = v / t. We need some guess at how long he jumps for.

The golden age Superman wasn't really quicker than the average person. He could run and move fast but he sort of jumped normal so let's estimate how long a person jumps for before their feet leave the ground. 1 second? No way. That's a pretty long time. 0.1 second? That seems a little short. Let's take the geometric mean and say 0.3 seconds. Then the acceleration is a = 75m/s / 0.3s = 250m/s^2. And then the force his legs exert is F=ma=100kg*250m/s^2 = 2.5*10^4N (N is newton's which is the unit for force or weight).

That seems big but what force does the average person exert? 1ft? Ya... It seems small but 1ft * 3 = 0.3m and an average person definitely can't jump 1m so let's just pick a number in between 50cm seems like a good guess. So the average person's potential energy at the top of their jump is PE = mgh= 100kg * 10m/s * 0.5m = 500J which means they need an initial velocity:
KE = 0.5 mv^2 = 500J ---> v^2 = 10m^2/s^2 ---> v=3m/s.

So the average person's acceleration is a = v/t = 3 / 0.3 = 10m/s^2. And the total force they jumped with is then F = ma = 100*10 = 1000N = 10^3N.

Superman's jump was 2.5 * 10^4N and and average person's was 10^3N. So Superman's jump is 25 times stronger than an average person's.

So to make Superman jump like a normal person on Krypton, the gravity must be 25 times greater. The gravitational acceleration on Krypton would then be 25 * 10m/s^2 = 250m/s^2. To get a gravitational acceleration that is 25 times greater on a planet the same size as earth, the mass of Krypton would have had to have been 25 times greater too. The mass of Earth is about 6*10^24kg which means the mass of Krypton would have been 1.5*10^26.

You should really read Physics of Superheroes to see what this means about what must have caused Krypton's destruction.

On that note, I think the Fermi Problem I pose tomorrow will be about Krypton's explosion.

What was the mass of Krypton?

Back in the dawning heyday of the superhero, writers were still determining the extent of Superman's powers. There was a time when Superman's powers didn't come from our yellow sun. One important distinction between then and now was that Clark couldn't fly. He was only "able to leap tall buildings in a single bound". It was explained that his super strength was a result of Krypton's larger gravity.

So, Detective, based only on the statement that he is "able to leap tall buildings in a single bound", what was the mass of Krypton? Some help for this Fermi question: It's a good idea to guess that Krypton is about the same size as Earth. You might be interested in remembering some high school physics that to jump to a height (h) you initially need the same kinetic energy (KE = 0.5 * m*v^2) as the potential energy at the top of your jump (PE = m*g*h).