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My question is about free fall from an airplane. Lets say your seat detaches from the plane and you follow a standard free fall.

$$v^2=v_0^2+2g\Delta x=\sqrt{0+2\cdot 9.8\,\mathrm{m/s^2}\cdot 10000\,\mathrm m}=443\,\mathrm{m/s}$$

Now, let's say that right before hitting the ground, you jump up from your seat (using the seat as a hard surface to jump from) to reduce the impact velocity of your body.

Here's were I'm lost: how would you calculate that? I was trying to calculate the force needed for your muscles to spring up, which should equal the force of gravity. In this case:

$$F_{muscle}= -mg= -100\,\mathrm{kg}*9.8\,\mathrm{m/s^2}= 980\,\mathrm N$$

Then what? In short, how much would the velocity be reduced by jumping up from the chair right before impact (assuming you can exert $980\,\mathrm N$ on the chair), if any at all?

Qmechanic
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hey
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1 Answers1

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This would be a center of mass reference frame problem. The center of mass of the system (you and the chair) isn't going to change until one of the objects strikes the ground. Any force that you apply to the chair is only going to change the momentum of the chair, and by Newton's 3rd Law, your momentum will change exactly the same magnitude in the opposite direction. The momentum in the COM reference frame will remain constant.

The change in your own momentum can be used to calculate your velocity in the COM frame, which then, using vector algebra is added to the velocity of the COM (in the earth frame). It's not going to make much difference.

If you are starting with a specific force that you apply to the chair, the the time of force application will determine the momentum transfer, and the the mass of the chair matters will determine the velocity.

Bill N
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