Hancock Analysis

Intro: This week, I chose to analyze Hancock’s super powers. Hancock is played by actor Will smith and appears in only one movie; “Hancock”. His powers include flying, super strength and bullet resistance. In the movie, there are numerous scenes with obvious physical flaws for a variety of reasons; of which I chose five in particular to analyze. See attached pictures for individual calculations.  

Problem 1: A runaway group of gang members is driving down the LA freeway in a Cadillac Escalade. Hancock flies into the back of the car and attempts to negotiate with the bandits. When they do not comply, he stomps his feet through the floor, planting them into the road surface and bringing the SUV to a stop in 4.5 seconds. During the skid, the SUV travels 4.5 Escalade car lengths, which is equivalent to 25.2 meters. The car weighs 2732kg with all four people in it (three bandits and Hancock). My goal was to find the force that Hancock had to exert on the ground opposite the direction of motion to bring the car to a stop in that period of time. I also wanted to find force felt on the mortal passengers over that period of time. After working out the calculations, I determined that he would have to exert a force of 21,710.3N over 4.5 seconds on the road. I also figured out that each passenger (not including Hancock) would experience about 124.5lbs of force on their bodies over that same time period.

Part two of this problem is after he stops the car, he picks it up and flies to some undetermined altitude. After some more negotiations, he still isn’t satisfied so he drops the car. He lets the car go in free fall for 5 seconds before catching it and bringing it to a stop in roughly half a second. The car is traveling 49 meters per second before being brought to a stop, and the passengers feel a force of 1542lbs when it does stop.

Problem 2: One of the worst physics problems in the movie is when Hancock stops a freight train dead in its tracks (pun intended), without even budging. The train goes from traveling 20 meters per second to 0 meters per second in an estimated 1-second. The movie actually does a decent job of portraying what would happen to the train because we can see the cars piling up and de-railing themselves behind the stopped locomotive as a result of conservation of momentum. Assuming Hancock was capable of stopping a 10 million pound train, there would have to have been an astronomically high coefficient of static friction between his shoes and the ground, otherwise despite his strength, he would slide backwards after being hit with the train. To calculate the coefficient of static friction, we have to know a few things about the scene first. We can estimate the train weighs about 10 million pounds because that’s the weight of an average half-mile long freight train. The train was going roughly 45 mph (20.11 meters per second). And finally we know Hancock weighs 100kg. We know the sum of the net forces equals mass times acceleration, and since Hancock doesn’t move when the train hits him, his acceleration is zero. So after rearranging the equation, we see that the force of the train equals the force of friction. The force of friction equals mew times the force of the normal, which is mass, times gravity. So we plug in the numbers we know and solve for mew (which is the coefficient of static friction). After doing the calculations, we find that the coefficient of friction will have to have been at least 93,078.6 which is unbelievably high (and as far as we are concerned, impossible). To put it in perspective, the coefficient of static friction of Velcro is 6.

Problem 3: Quite often in the movie, as Hancock is walking down the street, he takes off flying by jumping off the ground. In one particular scene, he jumps straight up and reaches the coulds in 1.5 seconds. Based on the type of clouds shown in the movie, we can assume he flies upwards of 10,000 feet before punching into the could layer. I wanted to calculate the force he exerts on the ground when he jumps that high in that amount of time. To do this, we have to know the mass of Hancock, the distance he traveled and the amount of time he did it in, all of which we have. Assuming that he applies the force to the ground almost instantly, we can represent that as 0.01 seconds. After doing the calculations, he shows that Hancock would have to apply a force of 40,000,000N to the ground to achieve that result.  

Problem 4: In a flashback, Hancock is shown throwing a beached Grey Whale back into the ocean. The only question in this scene is whether or not the whale would survive the Impulse of the throw by Hancock. To find out, we first have to calculate the impulse. We know the mass of the whale is 36287kg or roughly 80,000 lbs. However that’s all we know besides that we can estimate that the DT of the impulse is about half a second. However we can figure out the final velocity of the throw by determining how far the whale went. The whale is in the air for 5.5 seconds. We know initial velocity in the y-axis is the same as the final velocity in the y axis, and assuming the whale traveled in a parabolic motion, you can divide the DT by 2 and do a free fall problem from that. The reason we can do this is because at the top of the arc, the velocity in the y is zero so the only force acting on it is gravity and the initial velocity is zero and we can divide the time by 2 and know that it falls for 2.75 seconds before hitting the water. Once we have that value (22.05 meters per second as the final velocity before it hits the water), we can use trig to find the velocity in the x-axis. To use trig, we need the launch angle of the whale which can be estimated at 15 degrees after watching the clip several times. Once we know the velocities in the x and y axis, we use the Pythagorean theorem to find the overall initial velocity (40.25 meters per second). Finally, we use that in the impulse equation and find that Hancock exerted a force of 2921103N to the whale. And I’m not marine biologist but I don’t think whales are designed to withstand that kind of force.

Problem 5: In this final problem, Hancock is in a standoff with a gunman in a convenience store. He grabs a candy bar and challenges the gunman to a draw; his candy bar vs the gunman’s bullet. We see Hancock throw the candy bar with such force that it strikes the gunman and sends him flying backwards through the store window and onto the street. The goal here is to calculate the velocity of the candy bar and find the force on the gunman after he is hit with it. We can estimate that he flies backwards roughly 25 feet in 1.5 seconds. Figuring the mass of the gunman is about 70kg, which means the candy bar would have to be traveling at least 9567.4 meters per second. That happens to be 28 times the speed of sound. When the candy bar hits the gunman, it applies a force of 8,132.4N over an estimated time period of 0.05 seconds.