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. 

This week we watched Armageddon, which is a fantastic film despite the horribly inaccurate physics portrayed in the movie. NASA’s plan for saving the earth was flawed in several ways. If we ignore the fact that people could actually land a shuttle on the surface of an asteroid and then drill down into it and plant a nuclear weapon, then there is still one huge remaining flaw. Given the size of the asteroid, one nuclear weapon would not even come close to doing what it is shown to do in the movie. Even the largest nuclear weapon ever thought of (not even successfully built) would only split the asteroid and move each piece a few hundred meters in either direction before both smash into the surface of the earth.

My new plan is based off NASA’s plan from the movie, but with a few changes. I would still send astronauts up to the asteroid but instead they would drop 5 Tsar bombs and blow up the asteroid 4 hours after it passed the moon. Then I would send a second and third team, each to land on the surface of the individual pieces of asteroid, and on the surface, detonate more bombs after another 4 hours, causing the pieces to accelerate in the y-axis and allow them to cover enough distance in the y-axis in the remaining 2 hours of time before they smash into the earth.

In doing the calculations to figure out how much explosive power would be needed to propel each half fast enough to clear the earth in the remaining 2 hours, I realized the amount of force needed would be light-years beyond anything mankind is capable of generating with the technology we have today. I determined it would take the equivalent of 5.18x10^11 megatons of TNT, per piece. That would be more than 5.18 Billion Tsar bombs per half of asteroid.

Eraser Movie Scene

            This week we watched “Eraser” which in my opinion, suffered in comparison to Mission Impossible III. Eraser failed across the board in cinematic categories such as acting, screenwriting, and of course, accurate portrayal of scientific principles; more specifically, physics. In one of the last scenes of the movie; an epic battle with futuristic weapons takes place on the docs of the Baltimore Harbor. Arnold Schwarzenegger, also known as “The Eraser” is seen firing two weapons that shoot a projectile “close” to the speed of light. For a moment let’s suspend our disbelief that any such weapon would even exist and just focus on the physics.

            The problem that we are investigating is, if the henchmen that Schwarzenegger shoots go flying backwards, shouldn’t he too go flying backwards from the recoil of the gun?

            Since we know the projectile is traveling “close” to the speed of light, we can estimate that as being 3.0x10^7 meters per second. Since they don’t tell us how much the aluminum rounds weigh, I estimated it as 1 gram, or 0.001 kilograms per round. Beyond that, the only things we have to know for the problem are the weight of Schwarzenegger and the weight of the henchmen. After a quick Internet search, I found out that Arnold was roughly 115 kilograms in the movie, and I estimate that the average henchmen is around 65 kilograms. Knowing all of this, we can use conservation of momentum to find out what really should have happened in that scene.

            After performing the calculations, I found that if Schwarzenegger actually fired both rail guns at the exact same time, he would have been launched backwards at a velocity of over 413.8 meters per second. In addition, the henchmen that were shot would have been launched backwards at a rate of 461.5 meters per second. That kind of acceleration would be fatal for anyone, even the Eraser proving that physics is clearly not a required course for anyone intending to major in film related studies.

Problem I - Ethan Hunt needs to jump a gap in the bridge. We know he makes the jump, but the question is, what angle would he need to take off at in order to make the jump?

Known: We know that he does make the jump, and after a quick Google search, we found out his height is 5 feet 7 inches. 

Relevant Quantities: In order to solve for the angle, we need to do trigonometry, but to get there, we need values for the triangle; more specifically, the velocity in the x and y axis at takeoff. To solve for those, the values we need are…

Ø  The size of the gap in the bridge.

Ø  Ethan’s velocity right before he jumps.

Ø  Forces acting on him after takeoff, which since we’re ignoring air resistance, is only gravity.

After analyzing the jump several times in the movie, I was able to come to the conclusion that the gap is roughly three and a half “Tom Cruises” wide (or about 19 feet). In order to find his velocity, I went outside and did several sprints too see what top speed I could achieve. I’m a runner, so I’m in pretty good shape and after several tests I averaged 7 meters per second. Granted his character might be in slightly better shape than I am, we have to consider that he was also wearing pants and a jacket which would slow him down so were going to assume his initial velocity is 7 meters per second. And finally, we know gravity is -9.8 meters per second per second.

Since we know the initial velocity in the X-axis, and we know the distance he has to travel, we can calculate the time it takes to do so, which comes out to 0.83 seconds. We can then use this to figure out the time in the Y-Axis because the times will be identical. We then have to solve for the initial velocity in the Y axis and we do that by taking the 3 known quantities that we have; time,  acceleration due to gravity, and the change in distance in the Y-Axis which is 0 meters. We know the change in distance in the Y-Axis is 0 meters because the side of the bridge that he is landing on is identical to the height of the side that he is taking off on.

Putting those numbers together in the appropriate Kinematic Equation gives us 4.067 meters per second for the initial velocity. Since we now know the initial velocities in both the x and y axis, we can use trigonometry to solve for theta. In the calculator, you would do “the inverse tan function of the value of the side opposite of theta, divided by the value of the side adjacent to theta.

Assuming you entered that properly in the calculator and you’re in degrees mode, not radians, you should have gotten a value for theta being 30.16 degrees. That is the angle that Ethan Hunt has to jump at to just clear the gap.

Problem II – At the end of the film, Ethan Hunt needs to run one mile across town to get to the location where the “Rabbits Foot” is hiding. The question is, what speed does he average on his way there/is that speed even humanly possible?

Known: We know that the “Rabbits Foot” is one mile away because he tells us that in the movie when he is talking on the phone. If you time the run, starting with when he exits the window onto the rooftop, to when he arrives at the door of the other place, it is exactly 100 seconds.

Relevant Quantities: In order to solve for his average velocity on this run, all we need to know is the distance he had to travel, and the time it took him to get there.

Ø  About 1609 meters.

Ø  Exactly 100 seconds.

Knowing that Velocity is equal to the change in distance divided by the change in time, all we have to do is plug in for both those values and we get an average velocity of 16.098 meters per second. That’s about 36 miles per hour.

The highest human “foot” speed ever recorded was 27.79 miles per hour (12.4 meters per second); which was set by Usain Bolt  and recorded during a 100-meter dash in the Olympics. So we can conclude that either Tom Cruise is the fastest man in the world, or this scene is not physically, nor humanly possible!

Problem III – Ethan Hunt leaps off the top of a building running full speed. He is attached to a wire that when engaged, will swing him from one building to another at which point we will detach himself and fall down to the roof of the second building. The question is; is that daring stunt even remotely close to being physically possible?

Known: We know that the height of the first building is 226 meters, and the height of the second building that he needs to land on is only 162 meters.  We assume that Tom Cruise can once again reach his top speed of 7 meters per second when he goes to jump off the building. And finally, we know that the gap between the two buildings is 47.5 meters.  

Relevant Quantities: There are numerous relevant quantities in this problem because there are multiple steps that will go into solving whether it’s possible or not. The relevant quantities include…

Ø  The height of the first building; 226 meters.

Ø  The height of the second building; 162 meters.

Ø  The distance between them; 47.55 meters.

Ø  His initial velocity in the x- axis when he jumps; 7 meters per second.

Ø  Initial velocity in the y-axis when he jumps which we can assume is zero considering he pretty much runs straight right off the top of the building.

Ø  The time that he is in free fall (before the wire is engaged).

Ø  The length of the wire.

Ø  How high off the ground he is when the wire does engage and he starts the swing to the second building.

Ø  Wind resistance during his free fall (after a few calculations it came out to an average of 350 N of force which would reduce his acceleration to -8.1 meters per second per second.

Ø  The time he is physically “swinging” on the pendulum which is roughly 12.6 seconds.

Knowing all of that, you will be able to solve for the distance he travels in the x and y-axis, which comes out to 45.5 meters and -171.1 meters respectively. Using the Pythagorean theorem, we can solve for the length of the rope which comes out to 177.0 meters. Using trig we can solve for the angle of the rope with respect to vertical which comes out to 15 degrees, and we can deduce that at the instant that the wire engages, Ethan Hunt is roughly 54.9 meters above the ground because if he has traveled -171.1 meters vertically, and the building is originally 226 meters tall (basic subtraction).  

At this point, Ethan Hunt is swinging on the end of the wire like a pendulum, and how high he reaches on the other side will depend on his velocity, and the negative work done by air resistance along the way to slow him down.