A ball is thrown straight up with an initial speed of 9.90 m/s. How much time elapses until the bowling pin returns to the juggler's hand?

I suppose the “ball” and the “bowling pin” are the same thing.
Take g = 9.8 m/s²
Take all upward vectors to be positive.

Initial velocity, u = 9.9 m/s
Displacement, s = 0 m
Acceleration, a = -9.8 m/s²

s = ut + (1/2)at
0 = 9.9t + (1/2)(-9.8)t²
4.9t² - 9.9t = 0
t(4.9t - 9.9) = 0
t = 0 (rejected) or t = 9.9/4.9
Time taken, t = 2.02 s

Ball goes up till its velocity becomes 0.
Time for this is
t= 9.9/9.8=1.01s. The same time it takes to come down. So the total time is
T=2.02 s.

If I throw a BALL up I do not expect to find a PIN coming down.

One way to start is to realize that gravity is symmetrical.
If you throw it up at + 9.90 when it comes down it will have a velocity of - 9.90
a change of - 19.8 m/s

Now acceleration = change in velocity / change in time
a = v/t
so t= v/a = - 19.8 / - 9.8 ~ 2.0 s Note that unless you know gravity at your location you cannot be confident of a third significant figure.
ie 2.02 s is possibly overclaiming the accuracy

3.14 radians, pie are square cornbread are round.