Fast and Furious 7: The Skyscraper Jumps

      If you have seen any of the Fast and Furious movies, then you would know that they are filled with ridiculous stunts. If you have seen the whole franchise, though, then you have probably noticed that the group of thuggish street racers, inexplicably become somewhat of trained, super-agent-like, fighting machines. By the seventh movie in the series, of course, the crew basically goes from street racing for money, to becoming a super team, pulling off insanely ridiculous missions.
  In Fast and Furious 7, Dom( Vin Diesel) and his crew take a trip to Abu Dhabi to recover a speeddrive. The only problem is they have to steal it from a billionaire's skyscraper penthouse. They gain access to the billionaire's house by attending his party, but like always, things go wrong. A commotion of fist fighting, and gunfire errupt, and  Dom and Brian( Paul Walker), who are in desperate need for an escape plan, figure their only option is to drive out of the skyscraper into another. Using the billionaire's expensive, rare sports car (a Lykan Hypersport), they successfully jump the gap of the skyscrapers, however, the brakes in their car go out when they land, and they are forced to jump yet another gap into another skyscraper.
While this scene is super badass, there is always the question of it's plausibility. To find this out we need to calculate how fast they would have to be going to successfully jump the gaps between the buildings, in the amount of time they did, for both jumps.

PART 1:
The skyscraper they start out in is Tower 2 of the Etihad Towers, which is 305 m tall. Dom and Brian, however, are on a floor of the building that is about 3/4 high of the total height of the building which means they are about 228 m up. As they speed out of the side of the building, they fall vertically about 2 stories into the next building. Because the average height of one story of a building is 10ft, or 3 m, it is safe to assume that they fell a total of 6m downwards, with an acceleration due to gravity of -9.8m/s^2

Y-direction
yf=yi+Vy,i(t)+1/2(ay)t^2

• Yi- 228m
• Yf- 222m
• Vyi- 0
• Ay- -9.8m/s^2
• t- ?

To find out time, we must rearrange the equation to take the 

square root of 2(Yf(222m)-Yi(228m)/Ay(-9.8m/s^2) 

This come out to t equaling 1.1 seconds.

Using google maps, I found that the distance from Tower 2 to Tower 1, of the Etihad Towers, can be estimated at about 100ft apart, or 30m. With this info, we can figure out how fast the car would have to be going to reach the other building in the amount of time Dom and Brian did.

X-direction

• Xi- 0m
• Xf- 30m
• Vx,i- 0m/s
• Vx,f- ?
• Ax- 0m/s^2
• t- 1.1s

distance=speed*time

speed=distance/time

velocity= 30m/1.1s= 27m/s(60mph)

Dom and Brian would have had to be going at least 60mph to successfully jump the gap of the skyscrapers. "They didn't have enough time to reach that speed though." Since they were in the rare Lykan Hypersport, this scenario could actually be possible. With some research, I found out that the 3.7 litre, twin turbo-charged, Flat-six engine of this super car can go from 0mph to 62mph in just 2.8 seconds.

PART 2:
After they miraculously make it to the next skyscraper, their brakes give out and they are forced to jump another, even bigger, gap. Since their Yf  for the first part was 222m, that means that is now their Yi. This time, since the gap is longer and their is more time for vertical drop, they go down about 4 stories, which is about 12m.  Again, we must find the time


Y- direction

• Yi- 222m
• Yf- 210m
• Vyi- 0m/s
• Ay- -9.8m/s^2
• t- ?

To find out time, again, we must rearrange the equation to take the 

square root of 2(Yf(210m)-Yi(222m)/Ay(-9.8m/s^2) 

This come out to t equaling 1.5 seconds.

Using google maps, again, I found that the distance from Tower 1 to Tower 3 can be estimated at about 147ft apart, or 45m. With this info, we can figure out how fast the car would have to be going to successfully execute the second skyscraper jump.

X-direction

• Xi- 0m
• Xf- 45m
• Vx,i- 60m/s
• Vx,f- ?
• Ax- ?
• t- 1.5s

distance=speed*time

speed=distance/time

velocity= 45m/1.5s= 40m/s(89mph)

To find the acceleration, we can take the change in velocity, and divide it by the time.

40m/s-27m/s= 13m/s

Acceleration= 13m/s/1.5s= 8.66m/s^2

Given the crazy sports car they were in, Dom and Brian could have easily met the speeds and acceleration requirement needed to successfully complete their escape. As far as landing perfectly on all four tires however, not so sure. On the other hand, is this ridiculous stunt possible? Yes. Do I recommend it? Well, thats up to you.