Linear Differential Equations word problem I had this quiz on LDE and I wasn't sure how to do this problem...I know how to do LDE but I couldn't come up with the equation to get me started. Any ideas on how to do this?

 A: Here is a start. Recalling the Newton's second law

$$ F = m a = m\frac{dv}{dt}. $$

where $m$ is the mass and $a$ is the acceleration. Now, when you throw the rock vertically upward, the gravity $g$ and the resistance of the air $r$ will act downward. that results in the equation 
$$  m\frac{dv}{dt}= -g - r .$$
Now, you have a differential equation with initial condition $v(0)=20$.
Notes: 
i) The speed at maximum height is $0$.
ii) The distance $s(t)$ is related to velocity with the relation

$$ v(t)=\frac{ds(t)}{dt}. $$   

A: Here are some hints to get started.


*

*Let down be the negative direction, so that $g=-9.81$.

*Let $v$ be the velocity of the rock.

*Use Newton's Second Law $F=ma$ to derive a linear ODE which the velocity satisfies.

*Given: $m=1$ is given, $a=\frac{dv}{dt}$ and $v(0)=20$.

*What forces are acting on the rock after it has been released (Hint: gravity $mg$ and air resistance $-\dfrac{v^2}{10}$)? How strong are these forces?


Edit in response to comment:
The forces acting on the rock are assumed to only be gravity and air resistance (in the absence of additional information), so that $F=mg-\dfrac{v^2}{10}$.
This gives the IVP
$$
m\frac{dv}{dt}=mg-\frac{1}{10}v^2,\, v(0)=20.
$$
The solution of this problem models the velocity of the rock.
