Even the Hulk has to follow Isaac Newton’s 3 Laws of Motion.

Sure, he might be the strongest there is, but there’s no getting around Newton’s Laws – especially that last one, which can be a stickler if he’s trying to jump *up*.

In the latest issue of **Champions** from Marvel (grab it at your local comic shop or digitally here at Comixology), the ship in which our team of heroes – Cyclops, Ms. Marvel, Viv Vision, Hulk, Nova and Spider-Man – are travelling in is shot out of the sky over the ocean. While contemplating their options for reaching land, Spider-Man suggests that the Hulk could jump up, only to be quickly corrected by Cyclops who explains that if Hulk jumps up, he’ll sink the floating ship. But…would he, or was Cyclops just being overly cautious?

*Okay – one further clarification…this is the young Cyclops, pulled from his early days as an X-Man, the Miles Morales Spider-Man and the Amadeus Cho Hulk. The whole team is made of young Marvel heroes who are seeking to change the world, and learning that wanting to and actually changing the world are two different things. Back to science.*

This isn’t the first time the Hulk has had to consider moving against or within water in the series. In **Champions #1**, when rescuing a shipping container full of young girls, Viv, Nova and Hulk all realized he had to let the container sink all the way to the bottom of a harbor before he could push up against something and get the container onto dry land.

The lesson here – you can be as strong as the Hulk, but if you don’t have something to push against, all those muscles just don’t matter. That’s **Newton’s 3 ^{rd} Law of Motion** – for every action force, there is an equal and opposite reaction force. For the Hulk to jump, he has to push against the ground, and in return, the ground pushes back on him with a force that is equal to, and in the opposite direction of the Hulk’s push. In the case of jumping up from the ship, the Hulk would push down on it, the ship would push down on the water, and…well, as Cyclops said, our heroes would be swimming home.

But…*really*? If the Hulk did try to do a “Hulk jump” up from the ship, would it be enough to sink it? To figure that out, we’d need to know two things: 1) how hard the Hulk pushes *down* when he jumps *up* and, 2) the buoyancy of the ship. Let’s start with the buoyancy:

**Floaty Things**

The upward force exerted on an object floating in a fluid is equal to the density of the fluid multiplied by the volume of the fluid displaced (the amount of water the ship pushes out of the way), multiplied by the acceleration due to gravity (9.8 m/s^{2}), or:

F_{B} = ρVg

Our crew is in the ocean (density – *that thing that looks like a weird lowercase “p” up there is the Greek letter Rho, which stands for density* -1027 kg/m^{3 }), and gravity is 9.8 m/s^{2}, but as for the volume of the ship (in meters^{3})…who knows. It has enough buoyancy to hold four teenagers (Viv can fly) and a Hulk up, but making an estimate would be anyone’s guess. So let’s not right now.

Let’s talk about the Hulk’s jump instead.

**Jumpy Things**

A lot of this work has been done already online here and there, perhaps most notably by *Wired* blogger and Physics Professor, Rhett Allain who looked at “The Physics of the Hulk’s Jump” Allain did a great job of working through the numbers, so we’re going to borrow some of his work and approach for this investigation (yes, we know that this is a different Hulk, but the numbers do carry over pretty well).

To understand how much force the Hulk pushes down with when he jumps, we need to bring in the work-energy theorem which states that the work done on a system is equal to the energy change of the system. In other words, when the Hulk jumps and goes up, he gains gravitational potential energy. That energy is equal to the work done on the Hulk to get there, that is, the amount the surface the Hulk pushed against *pushed him back* when he jumped.

Some numbers – again, we’re going with Allain’s estimations:

The Hulk’s mass – 293 kg (yeah – that may seem a little low, but it’s a pretty reasonable 645 pounds based from Allain’s analysis of the Hulk’s body in comparison to other, human bodies).

Average jump height – 120 meters. That’s 400 feet, and we’re going with it as an okay estimate of an “average” jump for the Hulk.

There are a couple of ways to approach the physics of a vertical jump, and we’re going to go with Allain’s, even though the answers from a variety of approaches get values for the force that are all in the same neighborhood.

Think about a jump in slow motion – or get out of your chair and go through the motions. There’s the pre-jump crouch, and the push off. If we reduce the Hulk to a point located at his center of mass (just below the navel, near the top of the hip bones), we could follow that movement from the crouch to the height at push off.

In other words, and borrowing Allain’s image from *Wired* for a second, the difference between s_{1 }and s_{2} is it – that’s when the Hulk (or you, if you’re jumping along at home) is applying a push down on the ground. Following Newton’s 3^{rd} Law, if the Hulk pushes on the ground, the ground will push back. Just for that short bit of time – the time it takes to move the center of mass from s_{1} to s_{2} – that’s the push from the ground. The force of the push down equals the force of the push up, and that push up is responsible for the final vertical height of the Hulk, you, or your favorite basketball player.

And in this instance, using the Hulk’s center of mass, the Hulk’s final height, h, is equal to the height minus s_{2}, (h – s_{2}).

Back to our work-energy theorem. The total work done in the jumping portion is the work done by gravity (pulling down) added to the work done by the ground (pushing up). That work is equal to the change in energy experience by the Hulk, in other words:

W_{net} = W_{grav} + W_{ground} = ΔE

Where ΔE is the difference in energy between position 2 (the start of the jump) and position 3 (the Hulk’s final height), so,

W_{net} = W_{grav} + W_{ground} = E_{2} – E_{1}

Physics 101: work = force multiplied by displacement. So – the work being done by gravity will be negative (it’s pulling down), and the work being done by the ground will be the force of the ground pushing up multiplied by the difference in the Hulk’s center of mass between position 1 and position 2. In other words:

W_{grav} = -mg(s_{2} – s_{1})

W_{ground }= F(s_{2} – s_{1})

So,

W_{net} = -mg(s_{2} – s_{1}) + F(s_{2} – s_{1}) = E_{2} – E_{1} (1)

But since we’re looking at position 1 as our zero position, E_{1} = 0 (its height is zero, so its gravitational potential energy, mass x gravity x height would be equal to zero), so

W_{net} = -mg(s_{2} – s_{1}) + F(s_{2} – s_{1}) = E_{2} (2)

We’re trying to solve for F here, but we’ve still got E_{2}. That’s kind of easy – the energy the Hulk would have at position 2 is going to be all kinetic, since he will be moving. We’re also assuming minimal wind drag; therefore the conservation of energy is in full effect. In other words, the total amount of energy the Hulk has at position 2 is the same as the energy he’ll have at the top of his jump. The energy he’ll have at the top of his jump (when his velocity is zero)? All gravitational potential energy, or mass x gravity x height. Previously, we said the Hulk’s height would be h – s_{2}, so the Hulk’s energy at position two can be calculated by mg(h – s_{2}). So now we’ve got the Hulk’s energy at position 2.

Solving equation 2 for F then, we get:

And replacing for E_{2}…

Plugging in the values of (again, estimates on heights for s_{1} and s_{2} taken from Allain’s estimations):

s_{2 }= 1.4 meters

s_{1} = 0.56 meters

mass = 293 kg

h = 120 meters

gravity: 9.8 m/s^{2}

(and we’ll leave the math for you to try at home…)

F = 405415.3 Newtons – or 4.05 x 10^{5} N

That’s…well, a lot. Just to remind us – that’s the force that the Hulk pushes down with in a jump, and the force which (thanks Isaac Newton and your 3^{rd} Law of Motion!) the ground pushes back, against the Hulk, throwing him into the air, up to 120 meters. And that’s the average force with which the Hulk would push down (and the ground would push up) – so part of the jump would push down/up more, some less. But using the average is fine for us.

So – a long mathematical tangent aside, let’s get back to our original question. Had the Hulk jumped up to an average height while on the partially submerged ship in **Champions #4**, would he have swamped it, as Cyclops said?

Let’s go back to our buoyancy – the force keeping that ship floating is:

F_{B} = ρVg

With ρ = 1027 kg/m^{3}, and gravity at 9.8 m/s^{2}.

We also have to consider that we’ve got mass on the boat already – the Champions are standing on it. So…

Average mass of the four teenagers – 50 kg x 4 = 200 kg

Hulk: 293 kg

Total mass on the ship: 493 kg.

That’s mass, not force. Force downward is your weight, or F_{g}.

F_{g} = mass x gravity, so:

F_{g} = (493)(9.8) = 4831 N (already pushing down on the ship)

So – for the ship not the be swamped when the Hulk jumps off of it, our buoyancy force (F_{B}) would have to be significantly greater than 4.05 x 10^{5} N (again, an average) + 4831 N (about 4.1 x 10^{5}N). How can we figure this out, or at best, estimate it? Put 4.1 x 10^{5} N in for F_{B} and solve this thing for the volume of the vessel – V.

Rearranging, then:

Plugging in our values, we get V = 40.73 m^{3}. In other words, the ship could handle the weight of the Champions and Hulk’s push if it was displacing at least 40.73 m^{3} of water, that is, if the volume of the ship was around 40.73 m^{3}.

So how do we figure out how big that ship is, or at least get our heads around the volume? Comparisons.

Google is famous for asking difficult questions to its employment candidates. One such question: how many golf balls would fit inside a school bus? We’re going to go with an estimate of 960 ft^{3 }for the volume of the bus, which converts to about 27 m^{3}. So – a little under two school buses worth of space would give us our 40 m^{3} that the ship needed to be in order to support the Hulk’s jump.

Let’s go back to the comic – the ship in which the Champions are flying looks sizable from the outside, but in a Doctor Who-ish twist, it actually looks pretty cramped on the inside. It’s hard to make an estimate, but we’re going to go with the insides of the ship are way less than just under two school buses in volume.

So – what was the question again?

In order not to submerge when the Hulk jumps, the ship would need to be displacing at least 40.73 m^{3 }of water.

Ain’t no way that little ship is doing that. If the Hulk jumped, he’d push that ship (and the Champions) right under the water.

Final result – **Cyclops made the right call**. The Hulk could have leaped off the ship, but not without sinking his team.

Thanks Cyclops, thanks physics, and thanks Isaac Newton.

*By the way – Champions is written by Mark Waid, who knows a thing or two about science and isn’t afraid to show it. We’ve featured his other work with Daredevil and Invisible Woman and SHIELD. *

*And also – yes, the issue features Cyclops using his optic blasts as a means to push the Champions’ impromptu boat, like a motor. You know, for someone who was so worried about Newton’s 3rd law a few pages earlier, he doesn’t seem to notice that when he’s pushing against the water with his optic blasts, the water would be pushing back with the same force (enough to move the boat). What we’re saying here is neck strain. Lots.*