Answer
(a) The maximum height this rocket will reach above the launch pad is $646m$.
(b) $16.4s$ will elapse after engine failure before the rocket comes crashing down to the launch pad and it will be moving at a speed of $112\frac{m}{s}$ before crashing.
(c) See the graph below
Work Step by Step
Let the origin of our coordinate system be at the launch pad with the (+)ve y-direction upwards.
The motion of the rocket consist of two parts:
i) The motion of the rocket with upward acceleration of $2.25\frac{m}{s^2}$.
ii) The motion of the rocket after the engine fails with acceleration due to gravity.
Since both the parts have constant acceleration, we can apply the kinematics equation.
(a) Let us find the velocity of the rocket when the engine fails.
$2a_{y}(y-y_{0})=v_{f}^2-v_{i}^2$ , $v_{i}=0$
$=>v_{f}=\sqrt (2a_{y}(y-y_{0}))$
$=>v_{f}=\sqrt (2\times2.25\times525)=48.61\frac{m}{s}$
This velocity will be the initial velocity of the rocket in the 2nd part of the motion.
We now calculate the distance the rocket climbs after engine failure before coming to rest at max. height $(y-y_{0})$.
$2a_{y}(y-y_{0})=v_{f}^2-v_{i}^2$ , $v_{f}=0$
$=>y-y_{0}=\frac{-v_{i}^2}{2a_{y}}=\frac{-48.61^2}{-2\times9.80}=120.6m$
$y=525+120.6\approx646m$
The maximum height this rocket will reach above the launch pad is $646m$.
(b) Let us consider the 2nd part of the motion from engine failure to crashing.
$2a_{y}(y-y_{0})=v_{f}^2-v_{i}^2$
$v_{f}=+-\sqrt (2a_{y}(y-y_{0})+v_{i}^2)=+-\sqrt (-2\times9.80\times(0-525)+48.61^2)=-112.5\frac{m}{s}\approx-112\frac{m}{s}$
[(-)ve since the motion is downwards]
and $v_{f}=v{i}+a_{y}t$
$t=\frac{v_{f}-v_{i}}{a_{y}}=\frac{-112.5-48.61}{-9.80}=16.4s$
Thus, $16.4s$ will elapse after engine failure before the rocket comes crashing down to the launch pad and it will be moving at a speed of $112\frac{m}{s}$ before crashing.
(c) Let us calculate the time it takes for engine failure.
$t=\frac{v_{f}-v_{i}}{a_{y}}=\frac{48.61}{2.25}=21.6s$
The rocket crashes down to the launch pad $(21.6+16.4)s=38.0s$ after blast off.
i) We graphed $a_{y}-t$, using $a_{y}=2.25\frac{m}{s^2}$ from t=0 to t=21.6s and $a_{y}=-9.80\frac{m}{s^2}$ from t=21.6s to t=38.0s.
ii) We graphed $v_{y}-t$, using $v_{y}=v_{0y}+a_{y}t$ for the two parts of the motion of the rocket where $a_{y}=2.25\frac{m}{s^2}$ for the (+)ve slope and $a_{y}=-9.80\frac{m}{s^2}$ for the (-)ve slope.
iii) We graphed $y-t$, using $y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$ for the two parts of motion of the rocket where the sign of $a_{y}$ gives the curvature of y.
