2 Fast 2 Furious 2 Marginal

2 Fast 2 Furious was a sequel to the movie The Fast and Furious. Brian O’Conner (Paul Walker) and Roman Pearce (Tyrese Gibson) perform fast street racing as well as incredible stunts. Some insane stunts are even hard to believe, so I will be analyzing the physics behind a stunt scene in 2 Fast 2 Furious.

The scene I will be analyzing consists of Brian and Roman jumping a 1969 Chevy Camaro into a motor yacht. The two accelerate, launch off a jump, and then proceed through the air with unorthodox trajectory and land nearly perfect on the rear of the large vessel. I will be questioning the impulse of their sudden stop as they collide with the yacht and the correct distance they should have travelled given their velocity and estimated height.

A tremendous amount of impulse should have been exerted as this collision happened. Impulse is equivalent to force multiplied by the total time of the event taking place. (I=F(t2-t1)). The event of the car stopping was nearly instantaneous; it will be given an event time of .5 seconds. To find the force of the car, we take the total mass and multiply it by the total acceleration. A 1969 Camaro has a mass of 3,135 lbs and the driver and passenger both appear to be in-shape, adult males averaging 180 lbs each; this is a total of 3,495 lbs. The motor yacht was moving at cruising speed which is around 12 mph. The acceleration was 120 mph to 12 mph. This is a total deceleration of 108 mph. Converting the units to kg (Kilograms) and m/s (meters per second), we get 1585.3 Kg and 48.28 m/s. 1585.3 kg x 48.28 m/s =76,538.28 N (Newtons). Impulse equals force multiplied by total time. 76,538.28 N x .5 s = 38,269.14 N-s (Newton seconds).

The jump itself was questioned by Roman as Brian sped to 120 mph and launched both of them off of the ramp. Since Roman questioned the jump, I’ll crunch the numbers for him to see if his fear was justifiable. The ramp, that the Camaro jumped, is estimated to be three meters long and one meter high. Using trigonometry , three is our adjacent side (b) and one is our opposite side (a). To find the angle needed (A), we use the equation Tan^-1 (⅓). Our result is 18.43 degrees. We can then make a larger triangle using the total height of the car (5.19 m) and the takeoff angle (18.43 degrees). We find the angle needed for our calculation (B) by adding the takeoff angle (A) plus our ninety degree angle (C) (18.43+90=108.43). We then subtract the given number from 180 degrees, leaving us angle B; 71.57 degrees. Our opposite side(a) is 5.19 m, our angle A is 18.43 degrees, and our angle B is 71.57 degrees. By plugging this information into the equation (c=a sin C/ sin A) [5.19 sin (71.57)/ sin(18.43)] we result in the adjacent side length (b); 15.57 m. Since this is a triangle of only half of the distance the Camaro travelled, we multiply this number by two. Our final distance is 31.14 meters (102.16 ft).

A tremendous amount of impulse should have been exerted as this collision happened. Impulse is equivalent to force multiplied by the total time of the event taking place. (I=F(t2-t1)). The event of the car stopping was nearly instantaneous; it will be given an event time of .5 seconds. To find the force of the car, we take the total mass and multiply it by the total acceleration. A 1969 Camaro has a mass of 3,135 lbs and the driver and passenger both appear to be in-shape, adult males averaging 180 lbs each; this is a total of 3,495 lbs. The motor yacht was moving at cruising speed which is around 12 mph. The acceleration was 120 mph to 12 mph. This is a total deceleration of 108 mph. Converting the units to kg (Kilograms) and m/s (meters per second), we get 1585.3 Kg and 48.28 m/s. 1585.3 kg x 48.28 m/s =76,538.28 N (Newtons). Impulse equals force multiplied by total time. 76,538.28 N x .5 s = 38,269.14 N-s (Newton seconds).

The jump itself was questioned by Roman as Brian sped to 120 mph and launched both of them off of the ramp. Since Roman questioned the jump, I’ll crunch the numbers for him to see if his fear was justifiable. The ramp, that the Camaro jumped, is estimated to be three meters long and one meter high. Using trigonometry , three is our adjacent side (b) and one is our opposite side (a). To find the angle needed (A), we use the equation Tan^-1 (⅓). Our result is 18.43 degrees. We can then make a larger triangle using the total height of the car (5.19 m) and the takeoff angle (18.43 degrees). We find the angle needed for our calculation (B) by adding the takeoff angle (A) plus our ninety degree angle (C) (18.43+90=108.43). We then subtract the given number from 180 degrees, leaving us angle B; 71.57 degrees. Our opposite side(a) is 5.19 m, our angle A is 18.43 degrees, and our angle B is 71.57 degrees. By plugging this information into the equation (c=a sin C/ sin A) [5.19 sin (71.57)/ sin(18.43)] we result in the adjacent side length (b); 15.57 m. Since this is a triangle of only half of the distance the Camaro travelled, we multiply this number by two. Our final distance is 31.14 meters (102.16 ft).

In the cut of the car midair between the bank and the yacht, we have an accurate estimate of the distance between the car and vessel. A 1969 Chevy Camaro is 15.5 ft in length. There were exactly four car lengths between the end of the ramp (bank of the shore) and the rear of the yacht. Multiplying 4 car lengths by 15.5 ft (length of the Camaro) we have a total distance of 62 ft.

Taking into consideration that the yacht was moving away from the bank and Brian landed the Camaro on the top bridge of the boat, we’ve come to the conclusion that the scene is possible and the jump would have had a trajected landing on the yacht. Given Roman’s perspective, he seemed afraid when in fact Brian would have made the jump. But given the thought of jumping over 100 ft while travelling at a velocity of 120 mph, Roman had every right to be scared.