# How fast is Wolverine moving during his jump at the very end

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1. Dec 7, 2014

### mrspock

Or is there a way to calculate his speed? I know that since he's technically moving backwards, it's not the same speed as the train. The train's speed is about 300 mph.

2. Dec 7, 2014

### Danger

He's not moving at all. He's suspended in a harness in front of a greenscreen.

3. Dec 7, 2014

### mrspock

lol. i mean hypothetically, if this were real...i'm trying to do calculations for a project.

4. Dec 7, 2014

### Staff: Mentor

Take the length of his body, find out how many body lengths he's traveled over the train during the second or two in the air, and that should give you the distance he's traveled and the time, which you can use to get his speed compared to the train.

5. Dec 7, 2014

### Danger

Oh... okay...
Well, sorry, but you can't do it. The entire situation couldn't happen in reality.
Even if it could, though, you'd need far more data than is supplied, such as his mass, the wind speed independent of the train's motion (ie: what someone on the ground would feel), what his aerodynamic properties are, etc..
Edit: Hi, Drakkith. I hadn't thought of that as a "calculation" when reading the question, but it would work.

6. Dec 7, 2014

### mrspock

yeah of course it can't lol, my assignment is proving why it wouldn't work :p
what equations would i use though? i thought drag force would be a factor but i'm not sure how to do the calculations

7. Dec 7, 2014

### mrspock

oh that makes sense, thank you! :)

8. Dec 7, 2014

### Bandersnatch

You could find out what is the distance between the overhead line masts you can see him passing (on google maps, perhaps?), and measure the time it takes him to move between them. Then just use V=d/t to find his average speed relative to the ground. This is similar to Drakkith's suggestion, but should be easier to measure than counting the body lengths he's travelled, and gives you directly the speed relative to the ground rather than to the train.

Or, you could use some assumptions about his aerodynamic properties, the wind speed, the velocity gained from the jump, and solve the differential equation of motion with air drag to find out how fast he should be going.

9. Dec 7, 2014

### Danger

I've been a Marvel fan ever since the first time that I stumbled across a Spidey comic in my late teens. Wolvie almost immediately became my favourite character because we're both Albertans, both pilots, don't have qualms about killing, and share a basic attitude. Just for the hell of it, though... officially he is 5'3" tall and weighs 300 lbs.. The producers of the movies took a few liberties in hiring Hugh Jackman. (Don't get me wrong; he owns that role the same way that Downy does Tony Stark, and I wouldn't have it any other way. I was sold from the cage match in the first film.)

10. Dec 7, 2014

### mrspock

wait he's only 5'3"??

11. Dec 7, 2014

### Danger

12. Dec 7, 2014

### mrspock

So what I'm trying to prove is that he shouldn't have been able to fly forward for that long in a perfect straight line parallel to the train.
I'm thinking of using Δx = 1/2 (vi + vf) Δt to calculate how far he really should've jumped and how he should have fell early on.
It's not shown in the video, but he goes on to claw the other guy in the face, I believe.
Would this method work, or do I need to use different kinds of equations that factor in drag force and such?

13. Dec 7, 2014

### mrspock

How would you find the drag force? And would you subtract that from the average velocity?

14. Dec 7, 2014

### Danger

And again... that information is not available. You are on an impossible mission if you want to do it other than Drakkith's way.

15. Dec 7, 2014

### Bandersnatch

@Danger come on, we can always use the spherical cow approach. I think we can go through it step-by-step, and get at least an approximation if we treat it as a Fermi problem.

O.k., first of all, you need to be comfortable with the projectile motion (SUVAT equations and all that)

To be able to solve this analytically, we'll have to make the first assumption of there being no drag in the vertical direction. It's not a bad assumption, as the speeds and distances involved are tiny compared to the horizontal motion.

We'll have two equations then, one for horizontal and one for vertical motion.
$ma_x=-F_D$
and
$ma_y=-mg$
(minus means towards the back of the train and down respectively)

The second one basically gives you the SUVAT equations. Look them up, if you don't know them by heart.
Use the one for velocity to find out how long Huge Jacked-man stays in air. After all, gravity is the only force dragging him down.
You'll have to assume some initial velocity - find out what are typical velocities attainable by humans from rest. Assume he's jumping either at 45° to the horizontal, or straight up.

Get back to us with the answer (and how you got there), or if you'll get stuck.

We'll deal with the drag equation later.

16. Dec 7, 2014

### Danger

I haven't seen a spherical cow since my divorce.

17. Dec 7, 2014

### Staff: Mentor

That's his "official" height I guess. Not that it matters of course since no one draws him the same way and his height relative to other characters is always different.

18. Dec 7, 2014

### Danger

Except that it's always been well established that Cyclops is very tall (I haven't looked it up, but I think that he's 6'6") and referred to Wolvie as "Runt". James Marsden who played Cyke is 5'9 1/2" tall, whereas Hugh Jackman is 6'2". They had to do a lot of creative camera work in the movies to make it appear that Cyke was taller.
By the bye, just as filming began, Hugh strolled into a convenience store in Toronto to buy a snack or a pack of smokes or whatever and it somehow came out that he was "Logan". The guy behind the counter immediately said (because all Canucks are Marvel fans), "You'd better play him Canadian, right? If you don't play him Canadian, there'll be trouble." Hugh of course is an Aussie with a noticeable accent. He immediately took on a vocal coach to make sure that he presented himself on film with an Albertan sound.

19. Dec 8, 2014

### sophiecentaur

The terminal velocity of a skydiver in max-track position is about 120mph. The drag force will be about equal to the weight. At over twice that speed through the air, the force will be more than four times his weight (assuming a quadratic law). If he needs to match and exceed the speed of the train, the force would be even more and, once he's in the air, the drag work would soon slow him down.
No only is it nonsense, it's real nonsense. Even a Wizard would have a problem with that scenario.

20. Dec 8, 2014

### Bandersnatch

He doesn't, though. The whole idea is that slowing down w/r to the ground is speeding up w/r to the train (note the direction of travel). I don't see the issue here.