Could Spider-Man Catch MJ in the No Way Home Trailer?
It’s sort of a classic Spider-Man theme. He gets a girl friend, and then he has to jump to save her as she falls. Of course, in the comics the girl is Gwen Stacy. He catches her, but in the process of stopping, she does not survive. Is that what will happen in No Way Home but with MJ? Who knows. Here is the whole trailer.
OK, so now it’s time for some physics. There are a few questions to answer here, so I’m just going to get started.
Based on the trailer, when will Spider-Man reach MJ?
In the trailer, MJ falls — or maybe she’s pushed. Anyway, you can see her falling and then Spider-Man goes after her. There’s a great shot of both of them as they move towards the ground. Of course the scene cuts away before he reaches her — so I guess I will just have to wait to see the full movie.
Or — how about this? Let’s use video analysis to plot the the vertical position of both Spider-Man and MJ during this motion. I’m going to use Tracker Video Analysis (it’s my favorite video tool).
Since this part is clearly in slow motion, I don’t have the actual position-time values. But that never stopped me before. I’m going to make a rough guess that the distance between scaffolding levels is 6 feet (1.82 meters). I also need to account for the panning motion of the camera — but that’s fairly straightforward with Tracker. Here’s what that looks like.
This is the vertical position (in meters) versus time (in fake seconds — since I don’t actually know the time scale). Right away, you might see a problem. Both plots (for Spider-Man and MJ) look VERY linear. For a normal falling object (including humans and even MOST super heroes), the only force pulling on them is the downward gravitational force. This force can be modeled as the product of the object’s mass (m) and the gravitational field (g = 9.8 N/kg). Using this with Newton’s second law, we can find the acceleration of a falling object. I’m just going to write this in the y-direction so that it will be a scalar and not a vector equation.
For an object with a constant acceleration, it’s position as a function of time can be described with the following kinematic equation:
Plotting y vs. t for this kind of motion would produce a parabola, and not a straight line. So…I guess this movie isn’t real. Oh snap. Well, it’s not a big deal — it’s just a movie, it doesn’t have to be real. Trust me, I’m fine with this.
But since the motions are linear with time, then the slope of the position-time graph would be Δy/Δt which is also the velocity (in the y-direction). From the actual data, Spider-Man is moving downward with a speed of 2.57 m/s (not actual seconds) and MJ is moving at 2.42 m/s. At the beginning of this clip, Spider-Man is 1.99 meters away from MJ. Let’s say he has to be 0.75 meters away to be close enough to catch her. How long would this take (in fake seconds)?
Since I have two values (the position of Spider-Man and MJ), I will need two y-variables. I’m going to call them y_s (subscript) for Spider-Man and y_m for MJ. They both start at different initial positions with their own velocities. I can use the above kinematic equation and just set the acceleration equal to zero. Oh, just to make things easier I am going to put MJ’s initial position at y = 0 meters.
I want to find the time (t) when the difference in positions is some value (I will call it d = 0.75 m. Oh, notice that v_s is the magnitude of Spider-Man’s velocity — that’s why there is a negative sign in front of the velocity. Now I just need to subtract these equations and solve for t.
Putting in my values for the distances and times, I get a fake time of 8.27 fake seconds. Even though I don’t know the fake-to-real time conversion, I can still use this time to find out the final position of MJ when Spider-Man reaches her. Putting it into the MJ position equation, I get a drop of 20 meters (that’s around 65 feet). So, is that enough time?
I’m looking at the video again, and at first glance it looks like they are on the Statue of Liberty. The construction seems to be a project to add a giant shield to the statue??? I must be wrong. Well, the Statue of Liberty is around 300 feet tall, so even if they started half way up I think Spider-Man might have enough time to stop her from hitting the ground.
Of course, there’s still the question of why they are falling at a constant speed — hint: it’s not due to air resistance, but I will get to that soon.
What would a realistic fall and catch look like?
Let’s set this up such that both MJ and Spider-Man have the same downward acceleration (-g = -9.8 m/s²). In that case, the only way Spider-Man can catch up to MJ is if he “jumps” downward to give himself an initial downward velocity.
OK, so the question is how fast would he need to start moving down in order to catch MJ? The answer depends on some conditions. I’m going to make the following assumptions:
- Where and when did Spider-Man start his fall? I’m going to assume that he starts at the same place that MJ fell, but that he jumps after she has traveled 3 meters. I think a 3 m head start is VERY conservative. It could be MUCH larger.
- MJ can fall 20 meters before she must be caught. That will give enough time for Spider-Man to slow her down with a small enough acceleration such that she survives.
- I’m again going to go with the 0.75 meter range for the distance between MJ and Spider-Man so that he catches her.
Let’s say we write down a kinematic equation for both MJ and Spider-Man. Just to make things easier, I am going to say that at time t = 0 seconds, Spider-Man starts at y = 0 meters with his jumping velocity and MJ starts at y = -3 meters — also with some velocity.
If MJ started from rest, then her velocity after falling a distance of 3 meters (I’m going to call this distance d_f — the fall distance) can be found with the following kinematic equation. Note that since she is falling downward, d_f is negative (before you freak out about the square root of a negative number).
So, this makes it double difficult for Spider-Man. Not only does MJ start at some distance from him, she is also moving with a non-zero velocity. With a fall distance of 3 meters, she would already be going 7.67 m/s (17 mph). If Spider-Man wants a chance of making the catch, he’s going to have to start off with a velocity even greater than that.
Just to get a feel for what’s going on, let’s make a graph. Here is a plot of the position of both Spider-Man and MJ during this fall. I’m going to give Spider-Man an initial velocity that’s two times the value of MJ — so 15.3 m/s. Check it out.
In this case, he catches her — quite easily. It only takes a total fall distance of about 6.5 meters for them to have the same position — it’s even less than that if you go with the catch distance of 0.75 meters.
OK, then what is the minimum speed Spider-Man needs in order to catch her before she falls 20 meters. Let’s start with MJ. If she falls 20 meters, and starts from rest then we can solve for her falling time (she starts at y = 0 meters in this case so her final position will be -20 m).
This gives a time of 2.02 seconds. Now I can use that exact same formula to find out how long it takes her to fall 3 meters — this is the delay time for Spider-Man’s jump. That same equation gives a time of 0.78 seconds. This puts Spider-Man’s total time to get to 20 meters at 1.24 seconds.
Now that I have Spider-Man’s time, I can use that same kinematic equation to solve for his initial velocity.
Putting in all the values, this gives a starting velocity of 10.1 m/s (22.5 mph). I mean, that’s fast — but hey, it’s Spider-Man.
Oh, just for fun — if Spider-Man jumped straight up with that same initial velocity, he would rise 5.2 meters. Clearly, that’s higher than any human can jump — but remember:
Spider-Man, Spider-Man.
Does whatever a spider can.
There’s more to that song, but I don’t want to put the whole thing here.
What about air resistance?
Previously, I said that I could ignore the air resistance for this case. Yes, if you have an object falling there should be two forces acting on it. First, there is the downward pulling gravitational force. Second, an upward pushing (opposite direction as the velocity) force due to the interaction with the air.
As he moves down and increases in speed, this air resistance force would increase. Eventually, the magnitude of the air resistance force would be equal to the gravitational force. At that point, the net force is zero and the object would fall at a constant velocity (just like in the trailer). We call this velocity the terminal velocity — because you would stop speeding up. Yes, it’s exactly what happens to sky divers.
So…why not here? The answer is that they don’t fall for a long enough time to get to a speed with a significant air resistance. How far would they have to fall? Oh, I’m glad you asked — here is the answer.
Even with a fall of 200 meters — the air resistance is mostly negligible. So — I was right to leave it out.
