Hollywood vs. Physics: When Reality Strikes!

Hollywood is known for bending the rules of physics, and in some cases disregarding these fundamental laws entirely. Millions of people flood the movie theaters to watch the latest film and are astounded by the amazing scenes they see. The audience rarely contemplates the legitimacy of the stunts, and would rather watch in blissful ignorance. It comes as no surprise that the Fast and Furious movie series can be found in the ranks of physics-defying films. However, there may be some truth hidden in the wild car chases and death defying stunts.

    The fifth film in the speed-demon collection, Fast Five, is mystifying in its abuse of the laws that govern our universe. Viewers sit spellbound as the characters on screen are tossed from speeding trains, flip cars with fancy maneuvers, and walk away barely scathed. The most attention grabbing scene of Fast Five is perhaps the car jump scene, where Dominic Toretto and Brian O’Conner drive headlong off a cliff and jump into the water far below. Toretto and O’Conner emerge from the water soon after landing and both seem perfectly fine, but would the two speed demons really survive their headlong swan dive?

Bring forth the physics!

    The first step in determining if Toretto and O’Conner would survive the jump off the cliff is to calculate how much time it would take them to fall. If one assumes the situation to be a freefall problem, the initial velocity in the y direction is 0m/s. The acceleration in the y direction is gravity, so -9.8 m/s². The initial position in the y direction is 150m, which is supposedly an average sized cliff (the only average thing in this film). The final y position is 0m, in which case the water would be the final point. Using this information and the equation yf = yi + viyt + (1/2)ayt2 we can determine that the time it took Toretto and O’Conner to fall was 5.53s. This is not a lot of time, but if you have ever launched yourself off a cliff this may feel like forever. The equation below further explains this equation:

yf = yi + viyt + (1/2)ayt2

0m = 150m + (0m)t + (1/2)(-9.8m/s2)t2

(0m - 150m) / (-4.9m/s2) = t2

30.6s2 = t2

t = 5.53s

    Onward and outward! Next up is to figure out how far from the cliff the two landed. This will be determined by movement in the x direction. The acceleration in the x direction is 0m/s² (sadly no rocket boosters on this car... yet). The initial position is 0m. We can assume that the car is going the same velocity as the train because both Toretto and O’Conner match its pace while driving in the car, thus the initial velocity in the x direction is 60mph or 26.8m/s. If we take the time from the first step (5.53s) and the equation xf = xi + vixt + (1/2)axt2 we will find that the duo travelled 148.2m after they drove off the cliff (that is almost a football field and a half - off they go, soaring through the air). The equation below further elaborates on how to reach this answer:

xf = xi + vixt + 12axt2

xf = 0m + (26.8m/s)(5.53s) + 12(0m/s2)(5.53s)2

xf = 148.2m

    But where has the time gone? Once we have established the time and distance we must find the velocity of Toretto and O’Conner just before they hit the water (a.k.a. How fast they managed to go before plummeting into the cold water). The initial velocity in the y direction is 0m/s and the acceleration is -9.8m/s² once again. The initial y position is 0m and the final y position is 150m. Using this information and the equation vfy2 = viy2 + 2ay(yf - yi), we are able to discover that they would hit the water going 54.2m/s. That is about 120mph (I do not recommend driving this fast EVER, especially not off a cliff). The equation below sheds more light on the problem described above:

vfy2 = viy2 + 2ay(yf - yi)

vfy2 = (0m/s)2 + 2(-9.8m/s2)(0m - 150m)

vfy2 = 2,940 m2/s2

vfy = 54.2 m/s

At last we can get the answers we all really want to know! The final step of this convoluted Hollywood tale will tell us if the boys would really survive this dramatic stunt (once again folks - Do. Not. Try. This. At. Home.). We have already found how long, how far, and how fast they would go, but now let us discover with how much force Toretto and O’Conner would hit the water. We will focus on both men separately, and assume they are both an average weight of 88kg (I hope their fragile egos are not broken by being called average). Their initial velocity in the y direction is 0m/s, and their final velocity in the y direction is 54.2m/s according to the third step of this adventure. The change in time it takes for the impact is 2s, as it happens very quickly (duh). According to the equation F = pt, and p = mvf - mvi , the force both men experience separately would be 2,384.8N (yuuuup, that is going to hurt). The equation below explains this further:

I = p = Ft

F = pt and p = pf - pi = mvf - mvi

F = (mvf - mvi) / t

F = [(88kg)(54.2m/s) - (88kg)(0m/s)] / 2s

F = 2,384.8N

    The absurdity of Hollywood comes through once more in this outrageous stunt. The average human bone breaks under 3,000N of force, yet many bones have been known to break with only 2,000N of force. Both men would most likely survive the fall but would sustain numerous broken or fractured bones, especially the rib cage, and extensive bruising. It is highly unlikely that the two would resurface from the water unscathed and still be able to walk away. It is lucky for these two action packed daredevils that Hollywood, and especially the Fast and Furious series, does not obey the laws of physics. Although it is enjoyable to watch our heroes fight, jump, and do mind blowing tricks, we must remind ourselves that reality, and physics, would not agree. So the next time you feel the urge to follow in a superheroes footsteps or re-enact a crazy stunt, remember that physics is always there to bring you back to reality. One way or another.