Quadratic Drag Force: Solving Differently

[tryingtolearn1]

Homework Statement

There are two questions in my book that are almost the same but have two different answers.

The first question:
A particle is projected vertically upward in a constant gravitational field with an initial speed of $v0$. Show that if there is a retarding force proportional to the square of the speed, the speed of the particle when it returns to its initial position is where $$\frac{v_ov_T}{\sqrt{v_o^2+v_T^2}}$$ $v_T$ is the terminal speed.

Second question:
A baseball is thrown vertically up with speed v o and is subject to a quadratic drag with magnitude $f (v) = cv^2$ . Write down the equation of motion for the upward journey (measuring $y$ vertically up).

Relevant Equations

$F=ma$

But the answers to these two questions confuses me.

For the first question the answer goes like: $$m\ddot y = -mkv^2-mg$$ $$\therefore \ddot y = v\frac{dv}{dy}$$ $$\therefore v\frac{dv}{dy}= -kv^2-g$$

but for the answer to the second question we have: $$m\dot v = mg - cv^2.$$

Both questions are very similar but with different answers. Why is the first question solving it using $\frac{dv}{dy}\frac{dy}{dt}$ instead of just $\dot v$?

 

[haruspex]

The first equation for the first question is the equation of motion during ascent, and it matches the answer to the second question. But the first question requires you to go further.
It is not in general possible, or at least not easy, to solve drag equations to get position as a function of time. But in the first problem you do not need speed or position as a function of time, but speed as a function of position. Using $\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}$ makes it a lot easier.