Answer
(a) $3.8~m/s^2$
(b) $15~m/s$
Work Step by Step
The origin is on the groung and the positive direction is upward.
(a) The rock takes 4 s to reach the highest height (30 m). At this point, its velocity is zero.
$v=0$, $t=4~s$, $y=30~m$, $y_0=0$, $a=-g$
$v=v_0+at$
$0=v_0-g(4~s)$
$v_0=g(4~s)$
Use $x=x_0+v_0t+\frac{1}{2}at^2$, but make $x=y$. That is:
$y=y_0+v_0t+\frac{1}{2}at^2$
$30~m=0+g(4~s)(4~s)+\frac{1}{2}(-g)(4~s)^2$
$30~m=g(8~s^2)$
$g=\frac{30~m}{8~s^2}=3.8~m/s^2$
(b) $v_0=g(4~s)$
$v_0=(3.8~m/s^2)(4~s)$
$v_0=15~m/s$
