Motion of ball (air viscosity concerned)

Question

Suppose a ball of mass $m$ is thrown vertically upwards from the ground. I understand that the speed-time graph would be somewhat like a distorted parabola. But what about the velocity- time graph (considering air drag or viscosity)?
According to me it would attain a kind of terminal velocity while falling down. But I am unable to interpret it mathematically.
And sometimes you really need mathematical intuition to see what is happening. So can anyone make a brief mathematical interpretation of this?
Thank you.

Thirsty for concepts · Accepted Answer

Let, the viscous force drag,
$${F}={k}{v}$$ where ${k}$ is a constant and ${v}$ is the velocity at any instant. While moving up (upward acceleration is negative),

$${ma} = {mg} + {kv}$$
$${a}={g} + frac{kv}{m}$$
While moving down (downward acceleration is positive),

$${ma}={mg}-{kv}$$
$${a}={g} - frac{kv}{m}$$
From any of the two equations, it is clear that
$$-frac {dv}{dt} propto {v}$$
$$frac {dv}{v} propto {-dt}$$
$$int frac {dv}{v} = {n}int {-dt}$$
Where $n$ is a constant
$${ln v} = {-nt + c}$$
$$e^{ln v}= e^{-nt+c}$$
$${v} = e^{-nt+c}$$
Thus the graph will have an asymptote which represents the terminal velocity.

Motion of ball (air viscosity concerned)

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