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:

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

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:

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

**Hint:** I think the issue here is that the motion here is not symmetric. If it starts out with a speed $u$, it is not necessary that it will have the same speed when it reaches the bottom - because the acceleration is not the same in both the cases.

In the case without air resistance, it is valid to write $t=\frac{u}{g}+\frac{u}{g}$, because the particle goes from $u$ to $0$ and then $0$ to $u$, which doesn't happen in this case.

Try studying the problem by taking into account the distance the projectile travels - because the distance it travels up is always the distance it'll travel down.

## Best Answer

The force F is taken to be positive if it is upward and negative if it is downward. So -mg means that the gravitational force is downward. The term -kv means that, if the body is moving upward (positive v) the drag force is downward, and if the body is moving downward (negative v), the drag force is upward. This is completely consistent with our expectations.