AIM
Derive the equations of motion (hence of velocity and acceleration too) in each coordinate for the following problem.
SETTING
First of all, this question is very similar (in a sense) to this one Projectile Motion with Air Resistance and Wind. However my problem is way simpler than that. First of all the $z$ axis points upwards, the $x$ axis is horizontal, positive to the right and therefore $y$
I was thinking of a projectile of mass $m$, launched at a $\textit{speed}$ $V$ and at an angle $\theta$ from the horizontal. Now I want to take into account the force of gravity $\underline{F}_g = -mg\underline{k}$ (where $\underline{k}$ is the standard unit vector in the $z$ direction).
The drag proportional to the velocity (not the square!) $\underline{F}_d = -k\underline{v}_{pa}$, where I have written $\underline{v}_{pa}$ to emphasize that it is the velocity of the projectile with respect to the air, and $k$ is the drag coefficient.
Finally there is a wind blowing in the $\underline{j}$ direction with speed equals to 2, so the velocity of the wind is $V_w = 2\underline{j}$
MY TRIAL
x-coordinate (horizontal): The only force is the drag due to the initial velocity in the $x$ direction so: $\underline{F}_x = m\ddot{x} = -k\dot{x}$, which is the standard equation in the projectile motion with air resistance and hence I get: \begin{equation} \dot{x} = V\cos{\theta}e^{\frac{-kt}{m}} \,\,\,\, \text{and} \,\,\,\,\, x = \frac{mV\cos{\theta}}{k}(1-e^{\frac{-kt}{m}}) \end{equation}
z-coordinate (vertical): Here I have the force of gravity and the drag, hence $\underline{F}_z = m\ddot{z} = -mg\underline{k} -k\dot{z}$ and therefore I get the following: \begin{equation} \dot{z} = -\frac{mg}{k} + (V\sin{\theta} + \frac{mg}{k})e^{-\frac{kt}{m}} \end{equation} And also: \begin{equation} z = -\frac{mg}{k}t + \frac{m}{k}(V\sin{\theta} + \frac{mg}{k})(1 - e^{-\frac{kt}{m}}) \end{equation}
ISSUE HERE IS WHERE I NEED HELP I THINK
now for the $y$ coordinate I don't understand how to find the equations. Indeed I launch the projectile in the x-z plane, so the initial velocity in the $y$ direction should be $0$ right? I don't understand how to use the wind blowing in $2\underline{j}$. Should I say that since there is this wind, then the initial velocity of the projectile will be the same as the one of the wind? Or how should I do it?
I understand that the only force that will be acting n that direction will be the drag, opposing to the velocity given by the wind. Do you have any idea of how I should solve it?
Thank you
The system can be specified on the orthogonal axes easily if you give it the initial conditions in cartesian axis: $\vec{v}=v_{x_0}\vec{i}+v_{0_y}\vec{j}+v_{0_z}\vec{k}$. If the velocity of the wind is $\vec{w}=w_y\vec{j}$ we can express the equations in
$\dot{\vec{v}} = -\vec{g} - k(\vec{v}-\vec{w})$
For any axis and using the given wind:
$\vec{i}: \dot{v_x} = -kv_x$
$\vec{j}: \dot{v_y} = -k(v_y-w_y)$
$\vec{k}: \dot{v_z} = -g -kv_z $
These are three independent first order differential equations that you can solve for, the first two are fairly simple and the last one you can find for a solution in this post.Then you can integrate them again to obtain the position.
The y axis is actually the simplest to solve. The wind is going to drag slowly the the projectile out the $y=0$ path and eventually it will converge to the wind velocity.
$v_y=(v_{0_y}-w_y)e^{-kt}+w_y$
You can find the whole problem solution here.