#### Answer

(a) The rock's velocity is -24.4 m/s
(b) The rock is in the air for 4.5 seconds.

#### Work Step by Step

Note that upward is the positive direction.
(a) $v^2 = v_0^2+2ay$
$v = \sqrt{v_0^2+2ay}$
$v = \sqrt{(20~m/s)^2+(2)(-9.80~m/s^2)(-10~m)}$
$v = -24.4~m/s$
The rock's velocity is -24.4 m/s
(b) $v=v_0+at$
$t = \frac{v-v_0}{a}$
$t = \frac{(-24.4~m/s)-(20~m/s)}{-9.80~m/s^2}$
$t = 4.5~s$
The rock is in the air for 4.5 seconds.