Projectile Motion with Air Resistence and Wind

Question

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

Answer 1

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.