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Will Nathan Drake Make This Jump in the Uncharted Trailer?

You've played the video game, but now there’s a movie coming out based on Uncharted. One part of the trailer really got me interested—from a physics perspective. It shows a cargo plane with a long string of large boxes roped together and hanging out the back. The main character, Nathan Drake, clings onto this chain of boxes. (He’s played by Spider-Man star Tom Holland.) He climbs the string of boxes one by one until he reaches the one closest to the plane, then he jumps, making a leap toward the interior.

I have no idea why Drake is doing this, but it opens up a great physics question: Does he make it?

They don't actually show him getting into the plane, because showing the full action scene would violate the golden rule of movie trailers—to only give us a teaser. That's fine, I can figure out how this ends on my own.

Video Analysis

The first step is to get some data from the trailer, using a video analysis app like Tracker. (There are others, but that one is my favorite.) With video analysis, I can look at the location of an object (Drake, in this case) in each frame of the video to get his horizontal and vertical position. Since this trailer plays 24 frames per second, each frame can also give the time value for the motion of Drake. With that, I can make the following two plots showing his x-position and y-position as functions of time.

Taking a look at just the x-position graph, we can define the x-velocity as the change in x (normally we write that as Δx) divided by the change in time (Δt). But since it's a plot of x vs. t, Δx/Δt would be the slope of that line.

Fortunately, Tracker has an option to analyze the data and find a slope. This puts Drake’s horizontal velocity at 3.37 meters per second. Since it's a mostly straight line, it indicates that he has a constant horizontal velocity.

But should a jumper have a constant velocity in the horizontal direction? For now, to make things simpler, let's just ignore the fact that this jump is into a flying aircraft, which means there could be some air resistance force.

In this case, after Drake jumps off that final box, there is only one force acting on him: the downward-pulling gravitational force, which is equal to the product of his mass and the gravitational field, g. Since there aren't any forces in the horizontal direction, his horizontal acceleration is also equal to zero (from F-net = m*a). With a zero horizontal acceleration, there is a constant horizontal velocity, just as we would expect.

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Now let's look at his vertical motion. From the data, this also appears to have a constant vertical velocity with a value of 1.61 m/s. However, with a downward-pulling gravitational force, Drake should have a vertical acceleration of -9.8 meters per second per second (due to the gravitational field). This would make the y-position vs. time graph a parabola instead of a straight line. From a physics perspective, this isn't realistic. (Don't worry, it's just a movie, so it's not really an issue.)

Will He Make the Jump?

We are just going to have to work with what we have, even if it's not perfect real-world physics. I'm going to assume that Drake jumps off the box with an initial velocity of 3.37 m/s in the horizontal direction and 1.61 m/s in the vertical direction. His horizontal velocity will be constant since there are no horizontal forces acting on him. In the vertical direction, he will have a downward acceleration of -9.8 m/s2. We can deal with that.

In fact, there is the following kinematic equation that gives the final y-position (y2) as a function of time (t), starting velocity (vy1), and starting position (y1).

From the video, I know both his starting and ending y-positions (y1 = -0.45 m, y2 = 0 m). However, I don't know how long this y-motion will take. But that's fine. In physics, this would be a projectile motion problem. Here's a really useful trick: The vertical and horizontal motion can be treated as separate calculations except for one thing that they share—the time.

The time it takes for Drake to move in the vertical direction is the exact same time it takes for him to move horizontally. This means that I can use the horizontal motion to calculate the time, and then use that amount of time in the vertical motion to find his final vertical position.

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When Drake makes his jump, he needs to get up to a vertical position of zero meters; that’s the position of the ramp and where I set the origin. If this final value is less than zero meters, he lands below the airplane. And that would be bad.

Determining the horizontal motion isn't too difficult. Since he has a constant velocity, I can find his final horizontal position with the following equation:

Check this out: I know the starting x-position (x1 = 2.4 m) and the final x-position (x2 = 0 m), so I can use the x-velocity to solve for the time it takes to complete the jump. (He's moving to the left, so that's going to be negative 3.37 m/s.)

Notice that in the trailer we don't see the whole jump, but, if we did, it would take 0.71 seconds to reach the back ramp of the aircraft.

Now, I can use this time and plug it into the vertical kinematic equation. This gives a final y-position of negative 1.79 meters.

That’s lower than zero, so there’s nothing but air below him. And remember: That’s bad.

We’re not done yet, but it’s worth taking a second to wonder why he ends up even lower than he started. It’s because even though his initial velocity is in the positive (upward) direction, the jump takes so long that the gravitational force stops his upward motion and makes him move downward at a faster and faster rate.

What About the Moving Air?

When you stick your hand out the window of a moving car, you can feel something pushing back on you. This is the interaction between your hand and the air molecules surrounding the car—we call this air resistance. The amount of force you feel depends on the relative speed of your hand with respect to the air, and the size and shape of your hand. At high speeds, this air resistance force can be significant.

Let's say the aircraft has a flying speed of 120 mph—I like that value because it's the same as the terminal velocity of a human skydiver. When someone falls through the air for a while, the gravitational force causes them to increase in speed. But this increase in velocity also increases the upwards-pushing air resistance. At some point not too long after a jump, the upward air resistance force is equal to the downward gravitational force. This means the total force is zero and the diver no longer accelerates. Instead, now they move at a constant speed. We call that the terminal velocity. Of course, humans can still adjust their body and interact with the air to turn and maneuver—that’s why skydiving is still fun.

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What does this have to do with Nathan Drake? If he is moving horizontally with respect to the air, instead of downward like a skydiver, then the air resistance force would push back against him horizontally. At that speed, this air resistance would be about as strong as the gravitational force pulling him down. If he’s not holding onto anything, the air resistance would push him back, making him fall behind the moving plane very quickly. If he wants to jump against this air resistance force, it's going to be super difficult.

But it’s not as bad as you think. The cargo plane is also moving through the air—and its motion can cause some weird stuff to happen. Just think about the plane pushing the air out of the way during its flight. As the plane moves forward, all that air then has to rush back to fill in the spot behind it, where the plane was previously. This movement of air is called wake turbulence. It's possible that at that part of the jump, the turbulence from the plane could push Drake up and even towards the cargo ramp to make the jump. That could prevent him from landing too low and missing the ramp.

Honestly, I have this hunch that he's going to reach the plane. It's just a feeling, but I guess I will have to watch the movie to find out.


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