Answer
(a) $1.2~s$
(b) $0.61~s$
Work Step by Step
Let the origin be in the pitcher's hand with the positive direction upward.
The motion of free falling objects is symmetrical. So, the velocity of the glove when it returns to the pitcher has the same magnitude of the initial velocity, but with opposite direction, that is, $v=-v_0=-6.0~m/s$
(a)
$v_0=6~m/s$, $v=-6.0~m/s$, $a=-g=-9.81~m/s^2$
$v=v_0+at$
$-6.0~m/s=6.0~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=12.0~m/s$
$t=\frac{12.0~m/s}{(9.81~m/s^2)}=1.2~s$
You can also use $x=x_0+v_0t+\frac{1}{2}at^2$. The initial and final position are the same: the pitcher's hand. So, $x=x_0$ and, consequently, $x-x_0=0$. Now, we have:
$x-x_0=v_0t+\frac{1}{2}at^2$
$0=(6.0~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$
$t[(4.905~m/s^2)t-6.0~m/s]=0$
$t=0$ (the pitcher throws her glove)
or
$(4.905~m/s^2)t-6.0~m/s=0$
$t=\frac{6.0~m/s}{(4.905~m/s^2)}=1.2~s$ (the glove returns to the pitcher)
(b)
$v_0=6~m/s$, $v=0$, $a=-g=-9.81~m/s^2$
$v=v_0+at$
$0=6.0~m/s+(-9.81~m/s^2)t$
$(9.81~m/s^2)t=6.0~m/s$
$t=\frac{6.0~m/s}{(9.81~m/s^2)}=0.61~s$
