Answer
(a) The speed of the rock just before it hits the street is 32.7 m/s
(b) The time that elapses is 5.6 seconds.
Work Step by Step
Let up be the positive direction.
Let $y_0 = +30.0~m$
(a) Let $v$ be the speed of the rock just before it hits the street.
$v^2 = v_0^2 + 2a(y-y_0)$
$v = \sqrt{v_0^2 + 2a(y-y_0)}$
$v = \sqrt{(22.0~m/s)^2 + (2)(-9.80~m/s^2)(-30.0~m)}$
$v = 32.7~m/s$
The speed of the rock just before it hits the street is 32.7 m/s
(b) $t = \frac{v-v_0}{a} = \frac{-32.7~m/s-22.0~m/s}{-9.80~m/s^2}$
$t = 5.6~s$
The time that elapses is 5.6 seconds.
