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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!

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    $\begingroup$ Welcome to Physics Stack Exchange. Looking at other posts on this site you'll find that they do not use all caps titles, so I have edited the title here. Note also that we use mathjax for mathematical typesetting. Please edit your post to use it. $\endgroup$ – DanielSank Feb 21 '16 at 18:13
  • $\begingroup$ In the equation, the upward direction is taken to be positive in terms of displacement, velocity, and acceleration. So, the signs in the equation are correct. If the velocity is upward (positive), the drag force is downward (negative). $\endgroup$ – Chet Miller Feb 21 '16 at 18:21
  • $\begingroup$ Thank you DanielSank. I just don't know how ot use mathjax. $\endgroup$ – Student Feb 21 '16 at 18:23
  • $\begingroup$ Chester Miller, I understand that the upward direction is taken to be positive, and therefore the acceleration should be negative cause is poiting downwards. But I don't understand why also the air resistance is downward. I mean I imagine the air resistance to oppose to the acceleration of the free fall. $\endgroup$ – Student Feb 21 '16 at 18:25
  • $\begingroup$ Click on the link I gave you. Or do a Google search. How do you think the rest of us learned to use mathjax? :) $\endgroup$ – DanielSank Feb 22 '16 at 2:51
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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:

Parachute drop.

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:

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


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:

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

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