Friday, March 31, 2023

# How would Deadpool jump into a moving vehicle?

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If you asked people to describe a scene from a Deadpool movie, I bet most of them would choose the bridge ambush scene. It basically goes like this: Deadpool just hangs out and sits on the edge of a bridge viaduct over a freeway. He does things that make him happy, like drawing with colored pencils. But he also waits for a car full of bad guys to pass under the bridge. At the right moment, he jumps off the viaduct and crashes through the vehicle’s sunroof. The action combat sequence follows.

You know what comes next, right? I’m going to analyze the physics of this Deadpool move. I’m not going to spoil the scene. I’m just going to add some physical fun. Well at least I am will have fun. Let’s start.

Really, you can think of this move in two parts. The first part is the jump from the upper deck where it falls on the vehicle and hits it at the right time. The second part goes through the glass sunroof while missing the metal parts of the roof (I guess).

Jump at the right time.

So let’s say you see a car driving on a road below you. When should you get off the bridge and start your free fall? It’s actually a classic physics problem – and I love it. The best way to start a physics problem is to use a diagram.

We have two moving objects in this situation. Deadpool goes down and up in speed and the car moves horizontally with (I guess) a constant speed. The key to these two movements is the time. The time it takes Deadpool to fall a distance from the bridge must be the same time it takes the car to travel the horizontal distance. So let’s start with the fall of Deadpool.

Once it leaves the bridge, there is only one force acting on it – the downward pull of gravitational force. A net force on an object means that that object will accelerate. In general, the magnitude of the acceleration depends on the mass of the object (which would be Deadpool, in this case). But wait! Do you know what else depends on Deadpool’s mass? Gravitational force. By putting this force in this force-acceleration relationship (called Newton’s second law), I get: