Finding Equations of Motion A package is dropped from an aeroplane travelling horizontally at speed $U$ at time $t_0$ and height $z_0$. The package experiences acceleration due to gravity $ \boldsymbol{F_g} = $ $m$ $\boldsymbol g$ and the wind resistance proportional to its velocity $ \boldsymbol F_w$ $=-\alpha \boldsymbol v$.
Using Cartesian coordinates $(x,y)$ solve Newton's equations to find $\boldsymbol v$ $(t)$ $=(v_x(t), v_z(t))$. 
I am confused with what the question is asking me to do, since the object is experiencing the wind resistance in a direction proportional to the vector sum of the $x$ and $z$ directions, am I supposed to find the net force exerted on the object by adding the wind resistance and gravitational force vectors together? If so, what will this result tell me about $\boldsymbol v$$(t)$ ?
 A: Note: The solution below is long only because I'm being very detailed, for pedagogical reasons.

The net force acting on the package is the vector sum of the force of gravity and the wind resistance:
$$
\mathbf{F}_{\mbox{net}} = m\mathbf{g} - \alpha\mathbf{v}
$$
Newton's second law of motion then tells us that the acceleration of the package, $\mathbf{a}$, will satisfy the equation $m\mathbf{a} = \mathbf{F}_{\mbox{net}}$, or
$$
m\mathbf{a} = m\mathbf{g} - \alpha\mathbf{v}
$$
Since acceleration is the time-derivative of velocity, this amounts to a differential equation for the velocity vector:
$$
\frac{d\mathbf{v}}{dt} = \mathbf{g} - \frac{\alpha}{m}\,\mathbf{v}
$$
Suppose our coordinate system is oriented so that its origin is somewhere on the ground, with the $z$-axis pointing vertically upwards and the $x$-axis pointing horizontally along the same direction the airplane is flying.
In that coordinate system, the acceleration of gravity is represented by the vector
$$
\mathbf{g} = -g\,\hat{\mathbf{z}}
$$
(that is, gravity points downwards) while the velocity of the package is represented by the vector
$$
\mathbf{v}(t) = v_x(t)\,\hat{\mathbf{x}} + v_z(t)\,\hat{\mathbf{z}}
$$
where $\hat{\mathbf{x}}$ and $\hat{\mathbf{z}}$ are the unit vectors along the $x$ and $z$ axes, respectively.
Then, the differential equation above reduces to the pair of differential equations
$$
\frac{dv_x}{dt} = - \frac{\alpha}{m}\,v_x
$$
$$
\frac{dv_z}{dt} = - g - \frac{\alpha}{m}\,v_z
$$
Let's focus first on the equation for $v_x$. It can be rewritten as
$$
\frac{1}{v_x}\frac{dv_x}{dt} = - \frac{\alpha}{m}
$$
But 
$$
\frac{1}{v_x}\frac{dv_x}{dt} = \frac{d}{dt}\left(\ln v_x\right)
$$
so
$$
\frac{d}{dt}\left(\ln v_x\right) = - \frac{\alpha}{m}
$$
This is easy to solve now, and the general solution is
$$
v_x(t) = A\,\exp\left(-\frac{\alpha}{m}\,t\right)
$$
where $A$ is a constant. To find $A$ we must apply the so-called initial conditions of the problem. We're told that the plane is travelling horizontally with a speed $U$ at time $t_0$. Because the plane is travelling horizontally and the package is dropped (as opposed to thrown), we deduce that the horizontal and vertical components of the package's velocity, at time $t_0$, must be

$$
v_x(t_0) = U
\qquad\mbox{and}\qquad
v_y(t_0) = 0
$$

Thus,
$$
v_x(t_0) = A\,\exp\left(-\frac{\alpha}{m}\,t_0\right) = U
$$
so
$$
A = U\,\exp\left(+\frac{\alpha}{m}\,t_0\right)
$$
and

$$
v_x(t) = U\exp\left(-\frac{\alpha}{m}\,(t-t_0)\right)
\qquad (t \ge t_0)
$$

We now need to solve the differential equation for the vertical component of the velocity:
$$
\frac{dv_z}{dt} = - g - \frac{\alpha}{m}\,v_z
$$
This looks a lot like the differential equation for $v_x$ but it has the extra term, $-g$.
The theory of linear ordinary differential equations tells us that the general solution of a non-homogeneous equation such as this is equal to the sum of the general solution of its corresponding homogeneous equation and any non-zero particular solution of the non-homogeneous equation.
The homogeneous equation is
$$
\frac{dv_z}{dt} = - \frac{\alpha}{m}\,v_z
$$
and looks identical in form to the equation we just solved for $v_x(t)$, so its general solution is also
$$
v_z(t) = B\,\exp\left(-\frac{\alpha}{m}\,t\right)
$$
where $B$ is a constant. Now we need to find a particular solution to
$$
\frac{dv_z}{dt} = -g - \frac{\alpha}{m}\,v_z
$$
Any non-zero solution will do. Let's see if we can find a constant solution, i.e., $v_z(t) = V$ where $V$ is a non-zero constant. The derivative vanishes and we get
$$
0 = -g - \frac{\alpha}{m}\,V
$$
or
$$
V = -\frac{mg}{\alpha}
$$
Since $mg \ne 0$ (and $\alpha \ne 0$ as well, or else our solution would blow up!), $V \ne 0$ and we have found a satisfactory particular solution to the non-homogeneous equation. Thus, the general solution is
$$
v_z(t) = B\,\exp\left(-\frac{\alpha}{m}\,t\right) - \frac{mg}{\alpha}
$$
Now we need to apply the initial conditions again, this time for the vertical component. As mentioned above,
$$
v_z(t_0) = 0
$$
so
$$
0 = B\,\exp\left(-\frac{\alpha}{m}\,t_0\right) - \frac{mg}{\alpha}
$$
and
$$
B = \frac{mg}{\alpha}\,\exp\left(+\frac{\alpha}{m}\,t_0\right)
$$
Hence,

$$
v_z(t) = \frac{mg}{\alpha}\,\left[\,
\exp\left(-\frac{\alpha}{m}\,(t-t_0)\right) - 1
\,\right]
\qquad (t \ge t_0)
$$


Edit:
Note that
$$
\lim_{t\to+\infty} v_x(t) = 0
\qquad\mbox{and}\qquad
\lim_{t\to+\infty} v_y(t) = -\frac{mg}{\alpha}
$$
The physical interpretation of the above is that wind resistance is slowing the package down in both the horizontal and the vertical directions but, because of the effects of gravity, the slowing down in the vertical direction is compensated by the speeding up due to gravity and the vertical speed achieves what's known as terminal velocity.
Of course, the package isn't falling for an infinite amount of time so it is possible (for very weak wind resistance) for the package not to achieve terminal velocity and then hit the ground with a non-zero horizontal speed.
We can derive the terminal velocity more directly by noticing that, since the magnitude of the wind resistance increases with the magnitude of the velocity, the net force on the package will eventually be zero (if it doesn't hit the ground first):
$$
\mathbf{F}_{\mbox{net}} = m\mathbf{g} - \alpha\mathbf{v} = \mathbf{0}
\qquad\implies\qquad
\mathbf{v} = \frac{m\mathbf{g}}{\alpha}
$$
Since $\mathbf{g} = -g\,\hat{\mathbf{z}}$, we see that the terminal speed along the horizontal direction is zero and the terminal speed along the vertical direction is $-mg/\alpha$, as derived before.
