Answer
(a) $58~m$
(b) $34~m/s$
(c) $8.1~s$
Work Step by Step
(a) Let the positive direction be upward and the ground be the origin.
First, let's find the speed after the initial acceleration:
$v_0=0$, $\Delta x=26~m$, $a=12~m/s^2$
$v^2=v_0^2+2a\Delta x=0^2+2(12~m/s^2)(26~m)$
$v=\sqrt {2(12~m/s^2)(26~m)}=24.98~m/s$. This is the initial speed of the next step.
$v_0=24.98~m/s$, $v=0$ (at maximum height the velocity is zero), $a=-g=-9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$\Delta x=\frac{v^2-v_0^2}{2a}=\frac{0^2-(24.98~m/s)^2}{2(-9.81~m/s^2)}=32~m$
So, the rocket reached a height of $26~m+32~m=58~m$
(b) At maximum height the velocity is zero. Now, let the positive direction be downward and the origin at the maximum height:
$v_0=0$, $a=g=9.81~m/s^2$, $\Delta x=58~m$
$v^2=v_0^2+2a\Delta x=0^2+2(9.81~m/s^2)(58~m)$
$v=\sqrt {2(9.81~m/s^2)(58~m)}=34~m/s$
(c) Let $t_1$ be the time spent with the initial acceleration and $t_2$ the time spent in the free fall.
Let the positive direction be upward and the ground be the origin.
For $t_1$: $v_0=0$, $v=24.98~m/s$, $a=12~m/s^2$
$v=v_0+at$
$24.98~m/s=0+(12~m/s^2)t_1$
$t_1=\frac{24.98~m/s}{12~m/s^2}=2.08~s$
For $t_2$: $v_0=24.98~m/s$, $v=-34~m/s$, $a=-g=-9.81~m/s^2$
$-34~m/s=24.98~m/s+(-9.81~m/s^2)t_2$
$(9.81~m/s^2)t_2=24.98~m/s+34~m/s=58.98~m/s$
$t_2=\frac{58.98~m/s}{9.81~m/s^2}=6.01s$
$total~time=t_1+t_2=8.1~s$
