A body of mass $m$ falls under gravity and is retarded by a force proportional to its velocity. Write down the differential equation satisfied by the velocity $v(t)$ at time $t$.
Answer- $\frac{dv}{dt} = g-cv^2$ for some constant $c$
We have $$F = ma = m \frac{dv(t)}{dt} = mg- c v(t)$$ Where i am wrong?
On
You just consider a quadratic drag, which follows from Bernoulli theorem, $$F(v)=-\frac{AC_{d}\varrho}{2}v^{2}$$ So the equation is $$mv'=mg-\frac{AC_{d}\varrho}{2}v^{2}$$ By setting $c=\frac{AC_{d}\varrho}{2m}$ we obtain $$v'=g-cv^{2}$$ Which is separable and the solution is $$v(t)=\sqrt{\frac{g}{c}}\tanh(\sqrt{cg}(t-t_{0})+c_{1})$$ Where $c_{1}=\tanh^{-1}\sqrt{\frac{c}{g}}v(0)$
Your derivation is correct, indeed for a resistive force proportional to $v$, dividing by m your expression, we have
$$\frac{dv(t)}{dt} = g- \bar c v(t)$$
What is true is that the drag force for an object falling in the air is usually assumed proportional to $v^2$ but in general it depends upon the parameters for the motion (i.e. dimension of the object, viscosity of the fluid, speed, etc.) and in some context the assumption of proportionality to $v$ is true.