Assume that we throw a ball directly upwards with a velocity $v$ on the earth with downwards acceleration of gravity $g$ that has a quadratic drag force of $F_d = -bv^2$ acting on it. What would be the velocity of the ball when it comes back to the same position?
I'm asking on math stack exchange since I don't know how to formulate the differential equation nor what to do since there seems to be two moments: one when the ball reaches the top and then when it goes back down.
HINT
It helps to draw a free-body diagram to figure out the forces. Here it is easy, but still it is good practice. Let positive $y$ be pointing upward Newton's law is $$ m \ddot y = \sum F_i = - b \text{ sgn}(\dot y) \dot y^2 - mg, $$ along with initial conditions $y(0) = 0, y'(0) = v_0 > 0$, where sgn is the sign function. Can you finish from here?