Ludicrous Leg Man has never skipped a leg day, as a result of which he has infinitely strong and fast legs and all attendant secondary superpowers. Don't ask why, just roll with it.

While performing leg related actions such as running, jumping and kicking his body shrugs off all forms of damage, so he doesn't break his own bones, air resistance doesn't tear the skin from his legs, inertia doesn't push his brain into his spleen, compressive heating of air in front of him only burns off his clothes, and he can land perfectly unharmed (naturally in the typical hero pose).

While he can kick impressively fast his reaction times are still human, so when jumping he has to put all of his effort into one explosive movement (he can't cleverly run on air). He has a footprint (no pun intended) of 0.3m x 0.1m x 2legs = 0.06m2, and weighs about 90kg.

Given that he still has to deal with air resistance even if it doesn't kill him, he's jumping from sea level at STP and that the ground underneath him is not infinitely strong (assume it's granite, with a compressive strength of 220 MPa), can Ludicrous Leg Man overcome Earth's gravity (ie still be moving at escape velocity after clearing the atmosphere) without just digging his unusually manicured feet into the ground? If not, just how high can he get?

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    $\begingroup$ /only burns off his clothes, and he can land perfectly unharmed (naturally in the typical hero pose)/ ... wearing only smoldering rags. $\endgroup$
    – Willk
    Commented Dec 8, 2017 at 13:13
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    $\begingroup$ No-one(!) can overcome Earth's gravity. For reasons. Ludicrous Leg Man needs to achieve orbital velocity, not a certain altitude. Unless of course you are doing to postulate that he jumps so high he reaches the gravity well of another celectial body, like the Moon. Let me point out though that then he would have more power in his legs than any and all projectile weapons mankind has ever devised. With one possible, yet very unlikely, exception $\endgroup$
    – MichaelK
    Commented Dec 8, 2017 at 13:47
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    $\begingroup$ @MichaelK: I was actually referring to escape velocity (11.2 km/s for Earth). You’re right that he can’t escape gravity, but even in a universe containing nothing but earth you can go fast enough you never fall down again. $\endgroup$
    – Joe Bloggs
    Commented Dec 8, 2017 at 14:14
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    $\begingroup$ Let us assume that 1) Ludicrous Leg Man can crouch 1 meter and 2) as LLM jumps, the acceperation he experiences is not varying. To achieve 11 km/s on 1 meter, he has to achieve an acceleration of approximately 6 million g's. At 90 kg the force his feet would have to excert on the ground would be 54 GigaNewton. With a surface area of 0.06 square meters, the pressure is ~900 000 MPa. At a compressive strength of 220 MPa, that granite would be to his feet as floral foam is to a rifle bullet. $\endgroup$
    – MichaelK
    Commented Dec 8, 2017 at 14:32
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    $\begingroup$ @MichaelK: That's answer worthy information right there. The effects of such an impact would be pretty... impressive! $\endgroup$
    – Joe Bloggs
    Commented Dec 8, 2017 at 14:58

3 Answers 3


How much pressure can he put on the ground?

Since you were nice enough to give me the numbers you want, a 0.06 m$^2$ footprint can put $$220 \frac{\text{N}}{\text{m}^2} \cdot 0.06 \text{ m}^2 = 13 \text{ MN}$$ of force before shattering the ground or causing whatever other negative consequences.

How long is his foot in contact with the ground?

The next piece is trying to find out how long Ludicrous Leg-Man (LLM, for short) has his foot in contact with the ground, to determine how much work is done. Let us assume he starts from a crouch, and his center of gravity can go up by 1 meter before the force of his jump pulls him off the ground. The acceleration during his jump is calculated from $F = ma$ to be 146,667 m/s$^2$ (!!). Lets round that to 150 km(!!!!)/s$^2$. The relevant kinematics equation here is $$\begin{align}d &= \frac{1}{2} at^2 \\1 &= \frac{1}{2} \cdot 150 000 \text{ m/s}^2 \cdot t^2\\ t &= \sqrt{\frac{1 \text{ m}}{75000 \text{ m/s}^2}} = 0.00365 \text{s}.\end{align}$$

How fast is he launched?

Now we calculate the total speed after acceleration for that brief period of time: $$\begin{align}v_f &= v_i + at \\ v_f &= 0 + 150000 \text{m/s}^2 \cdot 0.00365 \text{s} = 548 \text{m/s}\end{align}$$

Now there are problems with this, specifically the shock waves created by surpassing the speed of sound. LLM is going to create a sonic boom. The instability caused by that sonic boom will probably make it really hard for him to jump where he wants to go. But that is complex modeling, and I'm going to ignore that for now. If you really want LLM to be Guile, ask Randall Munroe how that will go.

Also note, this is clearly not escape velocity.

How high can he go?

We first we can ignore air resistance and see. We set his initial kinetic energy from the launch equal to his potential energy at some height $h$ to get: $$\begin{align}\frac{1}{2}mv^2 &= mgh \\ \frac{1}{2}\cdot 547^2 \text{ m}^2\text{/s}^2 &= 9.81 \text{m/s}^2\cdot h\\ h &= 15290 \text{ m}\end{align}$$

A 15 km jump, not too bad! Nonetheless, even without air resistance, escaping Earth's gravity influence is not mildly feasible.

What about air resistance?

Thanks to my new favorite paper Calculation of Aerodynamic Drag of Human Being [sic] in Various Positions, we can estimate that the drag coefficient, $C_D$, for a person lying down is about 0.2. Of course LLM lying down in the air while going faster than the speed of sound is actually flying like Superman, so I think this is a good estimate.

This part of the math I don't have the space to do out, but I used a method pretty similar to what is seen here. First, we calculate terminal velocity as $$v_t = \frac{mg}{C_D} = 4414 \text{m/s}.$$ This is actually pretty high, based on our sleek aerodynamic super-flying profile and low $C_D$ value. Since terminal velocity is much higher than initial velocity, drag won't affect LLM that much. Assuming only vertical motion (i.e. LLM is jumping straight up) the equation for velocity as a function of time is
$$t= -\frac{v_t}{g}\log{\left(\frac{v_t+v}{v_t+v_0}\right)}.$$ Solving this for $v=0$ we get $t=52.6$, so LLM is in the air for 53 seconds at the top of his trajectory.

The equation for distance is obtained by solving the above for $v$ and integrating over time, to get $$z = \frac{v_t}{g}\left(v_t+v_0\right)\left(1-\exp{\left(\frac{-gt}{v_t}\right)}\right)-v_tt.$$ Plugging in a 52.6 second time, I solve this as 14096 meters, or 14 km. So, not that much different from our friction-less max, still plenty of juice to leap over mountains.

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    $\begingroup$ Awful thought inspired by MichaelK above: Can Ludicrous Leg man launch himself to orbit if he doesn't care about collateral damage? Could he kick the floor so hard the kinetic energy launches him upwards atop a pillar of flame? $\endgroup$
    – Joe Bloggs
    Commented Dec 8, 2017 at 16:12
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    $\begingroup$ @JoeBloggs Are you just going to keep adding questions in comments so I keep doing math all day? Because that would be great. $\endgroup$
    – kingledion
    Commented Dec 8, 2017 at 16:14
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    $\begingroup$ I can start a new question if you’d rather!! :-) $\endgroup$
    – Joe Bloggs
    Commented Dec 8, 2017 at 16:39
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    $\begingroup$ @JoeBloggs Actually please do. There are so many equations in that answer that making edits is straining my browser. This is a tough question, by the way, trying to figure out what happens to the ground when you jump like a ton of TNT. $\endgroup$
    – kingledion
    Commented Dec 8, 2017 at 16:40
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    $\begingroup$ "I mean, give a kick to a satellite. That will give you a boost. Also, it will anger NASA." -- dunno, NASA might have a satellite that needs deorbiting... $\endgroup$ Commented Dec 9, 2017 at 14:25

Plenty of math in kingledion's answer supporting his answer. Unfortunately, it's based on a false assumption--namely, that he can't jump higher than the rock under him can withstand.

This is false, he's going to be able jump considerably higher. I originally thought he could jump right off the planet but now I realize he can't--no matter how fast he jumps drag is going to bring him to a stop well before he leaves the atmosphere.

The thing is, so long as he jumps fast enough the strength of what he's standing on isn't the limiting factor. Rather, Newton's third law is at work. The rock under him is destroyed by the jump but it still has mass. It goes down, he goes up.

Lets suppose his legs go down at 70% of lightspeed. Draw lines downward from his feet, converging at a 45 degree angle. Any mass within that area has no way to escape (it would have to exceed lightspeed in order to do so) and thus must be pushed down.

I don't have the time to try to figure the trapped volume, but my gut says he ascends at at least 1% of lightspeed. However, no matter how fast his jump he's going to come to a stop when he's displaced as much atmosphere as his weight.

(Note: 70% was simply picked to make the 45 degree angle. Different velocities give different angles.)

(There is a way he can jump off the planet, though: Jump very, very hard. Drag brings him to a stop and he falls back to his jump site. Now, however, there is a huge shock wave pushing the air away from his location and the upward path from it. He jumps a second time, this time in near vacuum. While the problem says he can survive the jump no matter what, after he jumps there's going to be an incredibly destructive blast wave that will catch up with him. Does he survive the blast wave and radiant energy of a multi-gigaton blast?)

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    $\begingroup$ Where did the bit about air displacement come from? Reason I ask is because that statement doesn’t take momentum into account, which seems wrong. $\endgroup$
    – Joe Bloggs
    Commented Dec 9, 2017 at 9:45
  • $\begingroup$ I don't have enough rep to make a comment yet but i just wanted to reply to Loren Pecthel's answer where he states that he will stop once he displaces a mass of air equal to his own mass. This is not correct but is actually fairly close to the actual answer, that is he will stop once he imparts momentum to the surrounding air equal to his own momentum. Here is what i think is a rather interesting link to explain this: en.wikipedia.org/wiki/Impact_depth $\endgroup$ Commented Dec 9, 2017 at 14:09
  • $\begingroup$ A lot of materials are able to compress at least somewhat by having their atoms squished between each other though, so you can only consider the mass the path of his feet sweeps out. $\endgroup$ Commented Dec 9, 2017 at 14:41
  • $\begingroup$ @AJMansfield Of course the trapped mass gets tremendously compressed. Newton's third law doesn't care--he still goes up, it goes down. $\endgroup$ Commented Dec 9, 2017 at 17:30
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    $\begingroup$ @JoeBloggs Think of the energy involved. If his legs go down at 70%c the energy liberated will outshine the biggest H-bomb ever built. $\endgroup$ Commented Dec 9, 2017 at 17:32

Seeing as that LLM can jump infinitely fast with infinite power and basically can't be hurt, if he realy puts hit back into it he will reach light speed instantly and rip a hole in the planet under him through the opposite force.

The infinitely strong gravity waves caused by the inifinite spike in momentum will probably destroy the remainder of the planet, and, incidentally, the universe with it, losing him his frame of height measurement.

Be ware of infinities ;)

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    $\begingroup$ Actually, no. As his legs move down faster and faster there comes a point ever so slightly before lightspeed where the mass under his feet is compressed into a black hole. (Albeit a momentarily flat one) Now his feet get swallowed up rather than pushing down, acceleration ceases. Of course the black hole rapidly circularizes exposing new surface--but it likewise is compressed into a black hole. The process stops when his legs reach full extension and the black hole near-instantly evaporates. Legless man is sitting on top of a very big boom. $\endgroup$ Commented Dec 9, 2017 at 4:19
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    $\begingroup$ @LorenPechtel That's his final Heroic Sacrifice. $\endgroup$
    – wizzwizz4
    Commented Dec 9, 2017 at 11:17
  • $\begingroup$ @LorenPechtel the black hole (black plate?) There isn't a cut-off point where the hole suddenly collapses into itself and creates a void, let alone faster than light speed. It's just matter being compressed at the speed of his legs as he is pushing it down until it can't be compressed further. Since its still mass, action=reaction still applies and the black hole is pushed throught the planet as LLM goes up. $\endgroup$
    – durandal
    Commented Dec 9, 2017 at 12:35
  • $\begingroup$ Black hole? Is it not just really dense piece of granite? Black hole needs huge amounts of mass in addition to great density, I think.. $\endgroup$
    – diynevala
    Commented Dec 10, 2017 at 17:17
  • $\begingroup$ @diynevala "In principle, a black hole can have any mass equal to or above the Planck mass (about 22 micrograms)." en.wikipedia.org/wiki/… $\endgroup$
    – durandal
    Commented Dec 25, 2017 at 2:42

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