A common action movie stunt involves the hero jumping from a deck or a ledge dow

Question

A common action movie stunt involves the hero jumping from a deck or a ledge down into a moving convertible. Consider the following situation: James Bond (mass= 87Kg) and the enigmatic spy Jinx (mass=69Kg) are trying to stop a runaway convertible as it emerges from a parking garage by jumping into it from the deck immediately above the exit. Timing their leaps perfectly, and synchronized perfectly with each other, they drop into the seat of the moving car (mass=950Kg). The car is in neutral, and its speed just before they hit the seat is 10.4 m/s. When Bond and Jinx land in the car, it slows to 9.6 m/s. E) When they hit the seat, the back of the seat exerts a huge normal force on Bond and Jinx, which rapidly accelerates them up to the car's final speed even as the car slows down. Investigate the relationship between the force applied to the people and the distance that they sink into the seats. How far back will they have to sink into the seats (relative to the car) in order for the horizontal accelerating force exerted on them to be within a tolerable range? (You may want to do some web research on what might constitute a "tolerable range," but remember that we're talking stunt actors, not ordinary mortals.) The seats will have to be heavily padded on both the back and the bottom, and modified to slide back under the impact in order for this stunt to work. Based on your analysis, would it be possible to modify and equip a car so that the stunt could be performed by live actors (actually, ver good stunt doubles) in real life? Or is it so far beyond the realm of possiblity that you have to rely entirely on photographic trickery and/or computer animation?

Explanation / Answer

Total mass of Bond and Jinx = 87+69 = 156 kg

Mass of the moving car = 950 kg

Initial velocity of the car = 10.4 m/s

and final velocity of the car = 9.6 m/s which means that the car deccelerates.

The normal force = mg = (950+156)*9.8 = 10838.8 N

Total work done = m1v12/2-m2v22/2

= (950)*10.4 -(950+156)*(-9.6) = 20497.6 J

So, distance travelled = Work done/force = 20497.6/10838.8 = 1.891 m

To the high normal force, the peopel sink a distance of 1.891 m into the seats.

So, acceleration = 10838.8/(950-156) = 13.65 m/s^2 and for this they have to travel a distance of

10.4^2+9.6^2/2*13.65 = 7.377 m

which they have to sink into the seats (relative to the car) in order for the horizontal accelerating force exerted on them to be within a tolerable range.

The tolerable range of force = (950+156)*13.65 = 15096.9 N which is less than the normal forces acting on the actors. So, this stunt can be performed by live actors in real life.

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