How would Deadpool soar right into a transferring car?


If you asked People describing a scene from a Deadpool movie, I bet most of them would choose the bridge ambush scene. Basically it goes like this: Deadpool just hangs around and sits on the edge of a bridge over a highway. He does things that make him happy, like drawing with colored pencils. But he's also waiting for a car full of bad guys to pass under the bridge. At the right time, he jumps off the overpass and falls through the vehicle's sunroof. An action battle sequence follows.

You know what's next, right? I'm going to analyze the physics of this Deadpool train. I am not going to ruin the scene. I'm just going to add a bit of fun physics. At least I'll have fun. Let's begin.

Really, you can think of this step in two parts. The first part is the jump off the overpass bridge where it falls on the vehicle and hits it at just the right time. The second part goes through the sliding glass roof while the metal parts of the roof are missing (I assume).

Jump at the right time.

So let's say you see a car driving on a road below you. When should you leave the bridge and start your freefall? This is actually a classic physics problem – and I love it. The best way to start a physics problem is with a diagram.

Illustration: Rhett Allain

In this situation two objects are moving. Deadpool is moving down and gaining speed and the car is moving horizontally at (I assume) constant speed. The key to both of these movements is time. The time it takes Deadpool to fall one 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.

As soon as he leaves the bridge, only one force acts on him – the downward movement of gravitational force. A net force on an object means that the object is accelerating. In general, the magnitude of the acceleration depends on the mass of the object (in this case Deadpool). But wait! Do you know what else depends on the crowd of Deadpool? The force of gravity. If I put this force into this force-acceleration relationship (called Newton's second law), I get:

Illustration: Rhett Allain