Furious 7 is the 7th movie in the highly successful car-based Fast and the Furious movie series, and also the last film that Paul Walker did. There are lots of high octane action scenes stacked full of mathematical content.
Math Situation 1: Car Building Jump
In one of the most memorable scenes in Abu Dhabi, Dom (Vin Diesel) drives a high end sports car through the window of a skyscraper and manages to jump the car into the adjacent skyscraper.
Now, this is obviously a pretty crazy stunt. But is it even remotely possible? We can analyze some of the basic math behind the jump.
First of all, there are actually two jumps – we’ll just say they’re approximately the same and analyze only one of them.
It looks like the gap between the two buildings is approximately 50 metres.
It also appears that the car drops about 4 floors during the jump. These towers have very high floors so let’s say it’s 4 metres per floor:
Total drop = number of floors * floor height
Total drop = 4 * 4
Total drop = 16 m
If we ignore the effect of air resistance and other dynamics like the car pitching forwards, we can calculate how long it will take for the car to fall 16 metres:
Fall time = (2 * fall distance / gravity)^0.5
Fall time = (2 * 16 / 9.81)^0.5
Fall time = 1.81 s
That means the car has to have enough horizontal speed to cross that 50 metre gap in 1.81 seconds. Let’s work out how fast the car needs to be travelling to cross that distance:
Required car speed = distance covered / time
Required car speed = 50 / 1.81
Required car speed = 27.6 m/s
Since most car stats are given using km and seconds, we can convert the units:
Required car speed = 27.6 m/s / 1000 m/km * 3600 s / hr
Required car speed = 99 km/hr
So the car has to leave the building travelling at least 99 km/hr.
The car they’re travelling in is a Lykan HyperSport, a very high performance sports car. It has an acceleration capability of 0 to 100 km/hr in 2.8 seconds – one of the quickest road cars on the planet.
Under ideal conditions, we can calculate the time it would take to accelerate to the required 99 km/hr from a standing start (in the movie the car has some initial velocity for the second jump):
Time required = required speed / acceleration
Time required = 99 / 100 / 2.8
Time required = 2.77 seconds
That’s not a lot of time required, but there’s another factor – how far the car will travel in that time – is there enough room in the building to hit this speed before exiting!
Distance travelled = 0.5 * a * t^2
We’ll need the acceleration in units of m/s/s – which means we need to do a couple of conversions:
Car acceleration = 0 to 100 km/hr in 2.8 seconds
Car acceleration = 35.71 km/hr/s
Multiply by 1000 to get from km to m, and divide by 3600 to get from /hr to /s:
Now we have the right units, we can calculate the distance travelled:
Distance travelled = 0.5 * a * t^2
Distance travelled = 0.5 * 9.92 * 2.77^2
Distance travelled = 38.1 metres
38 metres isn’t very far, so (given all the big assumptions and simplifications) the fundamental math suggests that this stunt might be remotely possible.
Real Life Example – Sugar Glider
Sugar gliders are a small possum (marsupial) that jump between trees and use their membranes to glide long distances.
Sugar gliders have a different trajectory through the air to the car in the movie. The car is falling more and more rapidly through the air – it’s falling rather than gliding. A sugar glider will follow a relatively straight downwards path because it has what are in effect small wings. A typical glider will fall 1 metre for every 1.8 metres of horizontal distance travelled.
Let’s say a sugar glider is jumping from the top of a 20 metre tall tree and trying to make it to a tree 30 metres away (both trees are on flat ground). Will the glider make it to the other tree in the air or will it land on the ground?
Fall distance = horizontal distance / glide ratio
Fall distance = 30 / 1.8
Fall distance = 16.7 m
Since the sugar glider is starting 20 metres off the ground, if it loses 16.7 metres of altitude during the jump, it will be able to make it to the trunk of the other tree (just) in the air.