First, a little kinematics. Assuming he leaps just high enough to make it over, he’ll have zero vertical velocity at the top of his trajectory. We can then calculate the initial speed with which he must have left the ground:

V2 – V02 = 2aΔy,

where *V* is the vertical velocity at the top (zero), *V0* is the

vertical velocity upon launch, a is the acceleration due to gravity and

*Δy* is about six feet, or two meters. Solving for *V0* we get:

V = (2(9.8m/s2)(2m))1/2 = 6.3 m/s.

Now let’s consider that a typical jump involves about 0.2 seconds of

contact time with the ground. From this we can calculate Bourne’s upward

acceleration when he pushes off:

a = Δv/Δt = 6.3 m/s/0.2 s = 32 m/s2.

Applying Newton’s Second Law we can now estimate how much force he

will have to apply to do this. (And don’t forget Newton’s Third Law: However hard he pushes down into the ground, the ground will push back up with an equal force.)

Fnet = Fpush – mg = ma,

where *Fpush* is the force the ground exerts upward on Jason and his

relatively small motorcycle and *mg* is their combined weight. Let’s

assume that his futuristic bike is super-light at 100 kg, and he

himself is a trim but solid 80 kg.

Fpush = mg + ma = 180kg (9.8m/s2) + 180kg (32m/s2) = 7600 N . . . or *1,700 lbs*!

Not even the winner of the World’s Strongest Man contest has muscles capable of exerting a force of almost a ton. Think about it this way: To accomplish this stunt, Bourne must not only be capable of jumping six feet straight up, he has to be able to do it while holding onto a 200-pound motorcycle. Now imagine how strong Vin Diesel had to be in *XXX* to leap at least five times as high. This prompts the question: Why all the death-defying? With hops like that, Jason Bourne should consider a more easygoing (and lucrative) career in the NBA.—Adam Weiner