Free-body-diagram of a falling ball The free-body-diagram of a falling ball where air resistance is not negligible, is the drag force ever greater than the weight of the ball? I mean mustn't it as it is slowing down? But then how is the terminal velocity obtained? Does it mean then that the drag force decreases, so increases first, then decreases?
At least in my textbook they draw that the drag force on a falling ball is greater than that of its weight, how does this make sense? Perhaps let's look at the graph below, at t1, is the drag force greater than at t2? And I mean the gravitational force will always be on it the same from the instant it is released, right?

 A: I don't understand how you concluded that the ball is slowing down. It isn't. Its acceleration is decreasing. If it were slowing down, the velocity-time graph would curve downwards. Here it is flattening which shows that its acceleration is approaching $0$.
According to Newton's second law:
\begin{equation}
F_{net} = m a
\end{equation}
When the ball is falling down, it is experiencing two forces; the weight of the ball itself and the drag force. The drag force is proportional to the velocity, so after attaining a specific velocity called the terminal velocity, the weight equals the drag force, $F_{net}$ becomes $0$ and the ball stops accelerating, which results in the flattening of the velocity-time graph shown.
If the drag force were greater than the weight, the ball would experience a net upwards force and hence would keep on moving upwards instead of coming down.
A: For a sphere drag force = 6πnrv (stokes' law) {v=velocity} so in start v=0 so drag =0 so net force is downwards
so as ball comes down net force is downwards so velocity of increases in downward direction until for some v (drag force)=(gravity)
at this point net acceleration is 0 thus velocity is constant therefore drag force is constant and gravity is constant at every moment.
So from this point onward velocity is constant & known as terminal velocity.
So from starting only the velocity is increasing till terminal velocity and not that velocity first goes to maximum and then decreases
