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 forceacceleration relationship (called Newton's second law), I get:
This means that the free falling acceleration doesn't depend on the mass. Deadpool will move downward with an acceleration of g, where g is the local gravitational constant with a value of 9.8 N/kg (which is equivalent to 9.8 m/s^{2}). But since he will have a constant acceleration, I can use the following kinematic equation that gives a relationship between position, velocity, and time.
In this expression, y_{1} is the starting position and y_{2} is the final position. If I set the ground level to be y = 0 meters, then Deadpool will start at a position h (from the diagram) and end at zero meters. The v_{y1} is the starting velocity. Since he just steps off the bridge, this value will be zero m/s. Finally, I already put in the acceleration of negative g. With this, I can solve for the time it takes Deadpool to reach the ground.
Don't worry, I'm going to put in a value for the starting height (h) soon. However, let's just leave it as a mathematical expression for now. With this time, I can figure out the distance the car (OK, it's an SUV) needs to be from the bridge when Deadpool jumps. If the vehicle travels at a constant speed, then I can write a function of its position as the following (I'm assuming the point right below the bridge is at x = 0 meters).
Notice that horizontal velocity of the car has a negative value since it's moving towards the origin point below the bridge. If I put in my expression for the Deadpool fall time, I can solve for the distance of the car from the bridge when he jumps. Here's what I get.
You know what's great about not putting in any numerical values? I can make some rough estimates for the height of the bridge and the speed of the car—then I can easily change them if I feel like it. Let's do that. Honestly, I have no idea of the actual height of the overpass—since it's not a real overpass (that's the magic of movies). How about a value of 35 meters? That sounds nice. For the speed of the car, they actually show a shot of of the speedometer later in the fight with a speed of about 65 mph (29 m/s). Here is the calculation along with a plot of the position of both the vehicle and Deadpool.
That's actual real python code. If you click the pencil, you can go in and change the starting values. Oh, and what about this? Suppose that Deadpool jumps off the bridge and he is just 0.2 seconds late in his timing. Would he still make the jump? Nope. In that case he would miss his target spot by 2.9 meters. That pretty much would miss the whole SUV and look rather awkward. At least Deadpool is basically indestructible if he crashes onto the road.
Getting Through the Sunroof
Let's say that Deadpool hits the SUV at just the right spot. Will he be able to crash through the sunroof glass or will he hit the metal roof? For this, we need some assumptions.

How large is the sunroof opening? Let's go on the bigger side, with a length of about 70 centimeters. Technically, I also need the width of Deadpool to find the effective opening length. I'm just going to say the effective length is 40 cm (I'll call this x_{s}).

I will continue to use a vehicle speed (v_{c}) of a 29 m/s (65 mph).

The jump height will still be 35 meters. Yes, this important because it determines how fast Deadpool is moving (v_{D}) when he hits the roof.

Deadpool has a height (h_{D}) of 1.88 meters (6 feet and 2 inches). OK, actually that is the height of Ryan Reynolds (the actor who plays Deadpool).

One last assumption. Deadpool crashes through the sunroof in a standing position (his full body length) instead of something like a crouching position.
So, what's the big deal with the sunroof crashing? Suppose that the vehicle was stationary and Deadpool dropped down on top of it. He would smash the glass and end up inside the vehicle. Simple. But what if Deadpool is moving straight down and the SUV is moving horizontally? That means that Deadpool has to have his whole body move through the opening in the time that the vehicle moves forward. Here is a diagram.
The important time interval is the time the car moves over this effective opening size. Since the vehicle is moving at a constant velocity, the time will be:
For this same time, Deadpool has to move down a distance h_{D}. If he jumps from a height of 35 meters, I can find his downward velocity when he gets to the top of the vehicle—yes, I'm assuming the 35 meters is the height from bottom of the bridge to the top of the SUV. Since we are dealing with changing speed over a distance, I can use the following kinematic equation (he still has a vertical acceleration of g).
Just to be clear: Here the h is the drop height and v_{1} is the starting velocity (which would be zero in this case). Yes, it's also true that this is the velocity of Deadpool when his feet reach the top of the vehicle. Technically, he would still be increasing in speed as he moves through the sunroof, but the increase is fairly insignificant. With this, I can calculate the time it takes for Deadpool to move the required vertical distance. Here's what I get.
From the motion of the car, I get an available time interval of 0.014 seconds. Using Deadpool, it will take 0.072 seconds to pass through the opening. So … he's not going to make it. Oh, here are my calculations in case you want to change the estimations. Just click the pencil icon to see the code and change the values to whatever makes you happy.
If he wants to pass through the opening, there are two options. First, he could have a bigger sunroof—but that's something that is out of his control. The second option is to move faster when he gets to the SUV, but that means he needs to jump from a higher bridge. But how high? If I set the final Deadpool speed equal to the distance traveled (his height) divided by the time interval from the car, I can solve for the bridge height. Here is what I get.
Putting in my values for the speed of the vehicle and the effective size of the opening, I get a bridge height of 948 meters (more than 3,000 feet). Actually, this wouldn't work. This final speed is 136 m/s (304 mph) and that's faster than the terminal velocity of a falling human. He wouldn't get to that speed—ever.
OK, let'e figure this out. How could he get into the SUV through sunroof? Here are some options.

Change Deadpool's falling position. OK, let's be clear. Deadpool wants to be cool. He wants to look cool too It just seems like the vertical standing position just looks the best. But what if he pulled his knees up into a tuck position? Yes, this would decrease his height—but it would also decrease the effective length of the sunroof. I still don't think it would fix the problem.

Start with a nonzero velocity. It's very obvious from the scene that Deadpool just steps off the bridge so that he will have an initial velocity of zero m/s. But what if he pushes down to give an initial downward speed? I think this might work. I'll leave it up to you to calculate the speed and power needed to do this move.

Start with an initial horizontal velocity. Instead of just dropping down, what if he gets a running start in the direction of the vehicle? That would reduce the relative velocity between him and the SUV and increase the effective length of the sunroof. Of course this would also make the calculation of the impact point more difficult—so I will just leave that as a homework question for you.

Just forget about the glass. I mean, it's Deadpool—right? What if he hits the glass but gets knocked in the back by the rear edge of the sunroof and he just doesn't care? I think I like this option the best.
Wait. There's one more thing to consider. Suppose he makes it through the sunroof without hitting the metal roof. Now what? Well, his horizontal velocity is still zero and the SUV is moving at 65 mph (or something like that). So, he's stationary inside of a moving vehicle. From the reference frame inside the car, it would be like he is moving backwards at 65 mph. I would probably crash into the seats and then fly out the back window. That would look funny, but it's probably not what he intended.