The notation in the problem statement seems misleading to me.
We are given symbols $P_1,$ $P_2,$ $V_1,$ $V_2,$ $v_1,$ and $v_2,$
seemingly describing objects $1$ and $2$ in a completely symmetric fashion,
whereas the problem statement actually is not symmetric.
The problem statement implies that $P_1,$ $P_2,$ $V_1,$ $v_1,$ and $v_2$
are all known, with $v_1 = \lVert V_1\rVert.$
The unknown entity is $V_2$; we have the constraint that
$\lVert V_2\rVert = v_2,$ but as long as that constraint is satisfied,
we can choose the direction of $V_2$ freely.
The mathematics of finding a suitable direction for $V_2$ were thoroughly hashed out in the answers to Intersection of two moving objects.
There is an algebraic treatment
which concludes with an equation that the angle $\phi$ (the direction of vector $V_2$) must meet in order to make Object $2$ meet Object $1.$
There are at most two solutions of the equation, but in some cases a solution of the equation gives a value of $\phi$ that causes Object $2$ to move away from Object $1$ (so they will never meet), while in other cases there are two values of $\phi$ that would cause the objects to meet, although only one of those values can be the one that makes them meet in the minimum possible time.
I find a geometric approach gives me more insight on the problem;
in particular, once the geometry of the problem has been set up
in this way
(see this answer
or this answer)
it immediately gives us the equation
$$
v_2 \cos \phi = -v_1 \cos\theta,
$$
where $\phi$ is the angle between $V_2$ and the line $P_2P_1$
and $\theta$ is the angle between $V_1$ and the line $P_2P_1,$
measuring both angles counterclockwise.
(The algebraic treatment comes to an equivalent conclusion.)
Another way to say this is:
In order for the two objects to meet, the component of $V_2$ perpendicular to the line between the objects must exactly cancel the component of $V_1$ perpendicular to the line between the objects.
This condition is necessary but not sufficient to solve the problem.
A condition that is both necessary and sufficient is:
The vector $V_2 - V_1$ must point in the direction from $P_2$ to $P_1.$
Geometrically, we can construct representations of all possible vectors of the form $V_2 - V_1$ (given that $V_1$ is fixed and the magnitude but not the direction of $V_2$ is fixed)
by drawing the vector $-V_1$ and then drawing a circle of radius $v_2$
around the tip of $V_1.$
We can then represent $V_2 - V_1$ by placing its tail at the tail of
$-V_1$ and its head at any point on the circle; the direction from the center of the circle to the chosen point is the direction of $V_2.$
The diagram in this answer
shows the case where $v_1 > v_2$ but where it is still possible to make the objects meet.
In this case the circle intersects the line $P_1P_2$ in two places,
giving two possible choices of a vector $V_2$ that will cause the objects to meet.
The choice that causes the objects to meet in the minimum possible time is the choice for which the magnitude of $V_2 - V_1$ is larger.
(In the diagram in the linked answer, that vector is $CG.$)
When $v_1 > v_2,$ it can also happen that there is no solution to the problem; object $2$ can only "catch" object $1$ if object $1$ is initially reducing its distance to object $2$ quickly enough.
When $v_2 \geq v_1,$ there is always a solution but only one solution
for $V_2$ that allows object $2$ to "catch" object $1,$
so of course that is also the minimum-time solution.
The solution is shown graphically in this answer.
The problem has nothing to do with the formula.
Given that the only force acting the object is air resistance which is proportional to its velocity, you simply have
$$\frac{dv}{dt}=-av$$
Integrate the ode to get,
$$\ln v(t)=-at+C$$
Use the initial condition $v(0) = 50$ to obtain the constant $C$,
$$C=\ln 50$$
Then, use the information at $t=10$, i.e. $v(10) = 40$ to obtain the equation below for resistance constant $a$,
$$\ln 40 = -10a + \ln 50$$
which yields,
$$ a = \frac{\ln 5 - \ln 4}{10}$$
Therefore, the differential equation for the situation is just,
$$\frac{dv}{dt}= \frac{\ln (4/5)}{10}v$$
Best Answer
By Newton's second law (rate of change of momentum is equal to net force applied),
$$m \frac{dv}{dt} = mg - \alpha v$$; $\alpha $ is a constant, $\alpha v$ is the resisitive force.
Integrate w.r.t. time to get $v$ in terms of $t$.
$v|_{t = 0} = 0$ and $v|_{t \to \infty} = v_t$, terminal velocity is attained when $\frac{dv}{dt} = 0$, i.e. $v_t = \frac{mg}{\alpha}$
$m \int_{0}^{\frac{v_t}{2}} \frac{dv}{mg - \alpha v} = \int_{0}^{t_0}dt\quad$; $t_0$ is the required time.