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 is just hanging out and sitting on the edge of a bridge overpass over a highway. He’s doing stuff that makes him happy, like drawing with crayons. But he is also waiting for a car full of bad guys to pass under the bridge. At just the right time, he jumps off the overpass and crashes through the vehicle’s sunroof. Action fighting sequence follows.
You know what comes next, right? I’m going to analyze the physics of this Deadpool move. I’m not going to ruin the scene. I’m going to just add some physics fun. Well, at least I’m going to have fun. Let’s get started.
Really, you can think of this move in two parts. The first part is the jump from the overpass bridge where he falls down to the vehicle and hits it at just the right time. The second part is passing through the glass sunroof while missing the metal parts of the roof (I assume).
Jumping at the Right Time.
So, let’s say you see a car driving along a road below you. At what point should you step off the bridge and begin your free fall? This is actually a classic physics problem—and I love it. The best way to start a physics problem is with a diagram.
We have two objects moving in this situation. Deadpool is moving down and increasing in velocity and the car is moving horizontally with (I assume) a constant velocity. The key for these two motions 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 Deadpool’s fall.
Once he leaves the bridge, there is only one force acting on him—the downward pull of the gravitational force. A net force on an object means 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! You know what else depends on the mass of Deadpool? The gravitational force. Putting this force into this force-acceleration relationship (called Newton’s second law), I get: