Parachute jumping with air resistance

Question

I need to write down a model for a man parachuting from a plane at a height $h$ above the ground, having a velocity $v_{\text{plane}}$. I've had a look at many models online and they all start saying that $F=ma=-kv-mg$. Where $k$ is the air resistance before the deployment of the parachute, $v$ is the velocity, $m$ is the mass of the man and $g$ is the acceleration. However, I think that the trajectory should also be in three dimensions, i.e. $F=(F_x,F_y,F_z)$, cause with the jump from the plane, the man should have an initial velocity coming from leaning out of the plane, and therefore there should also be some air resistance in that direction.

Is that correct or can we neglect it?

Furthermore, I don't understand why $ma=-kv-mg$, I mean this two forces should oppose eachother shouldn't they? The air resistance should be in the opposite direction compared to the acceleration due to the gravity. I really hope you can help me with this!

Answer 1

The air resistance should be in the opposite direction compared to the acceleration due to the gravity. I really hope you can help me with this!

No. The drag force simply points in the opposite direction of the velocity vector.

Now consider the following simplified model:

Assume the dropping plane was flying horizontally and parallel to the $x$-axis, at speed $v_0$, then at the drop point ($t=0$) the parachute has two velocity vectors with scalars:

$$v_x=v_0$$ $$v_y=0$$

As the chute doesn't deploy immediately, in the $y$ direction two forces act: gravity and air drag, so with Newton we can write:

$$ma=mg-\frac12 \rho C_{y,1}A_{y,1}v_{y}^2$$

Set: $\frac12 \rho C_{y,1}A_{y,1}=\alpha_1$

Then after integration between $t=0, v_y=0$ and $t, v_y$:

$$\large{v_y(t)=\sqrt{\frac{mg}{\alpha_1}\big(1-e^{-\frac{2\alpha_1t}{m}}\big)}}$$

Assume the chute opens at $t=\tau$ then for $t>\tau$ we can derive also:

$$\large{v_y(t)=\sqrt{\frac{1}{\alpha2}\big(mg-\big(mg-\alpha_2v_{y,\tau}^2)e^{-\frac{2\alpha_2t}{m}}\big)}}$$

With:

$$\large{v_{y,\tau}=\sqrt{\frac{mg}{\alpha_1}\big(1-e^{-\frac{2\alpha_1\tau}{m}}\big)}}$$

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The parachute also experiences drag in the $x$-direction. Prior to deployment of the chute (and assuming no side wind):

$$ma=-\frac12 \rho C_{x,1}A_{x,1}v_{x}^2$$

Or:

$$a=-\alpha_3v_x^2$$

$\frac12 \rho C_{x,1}A_{x,1}=\alpha_3$

On integrating between $t=0, v_x=v_0$ and $t, v_x$

$$v_x(t)=\frac{v_0}{1+v_0\alpha_3t}$$

And for $t>\tau$:

$$v_x(t)=\frac{v_{x,\tau}}{1+v_{x,\tau}\alpha_4t}$$

where:

$$v_{x,\tau}=\frac{v_0}{1+v_0\alpha_3\tau}$$