91Ó°ÊÓ

Magnitude of acceleration of the rocket, $a=30m/s^2$

Time of acceleration, $t=30s.$

The rocket reached at a height $h_1$in first $30s$which can be determined by,

$$\begin{gathered} h_1=\frac{1}{2}a{t}^2 \\ {h}_1=\frac{1}{2}×30×{30}^2 \\ {h}_1=13500m \end{gathered}$$

Reaching the height $h_1$the rocket gains a constant speed of

$$\begin{gathered} v=at \\ v=\left(30×30\right)m/s \\ v=900m/s \end{gathered}$$

The ascending time of the rocket can be determined by the equation $t\text{'}=\frac{v}{g}$as,

$$\begin{gathered} t\text{'}=\frac{900}{9.8}s \\ t\text{'}=91.84s \end{gathered}$$

The distance covered by the rocket during ascending,

$$\begin{gathered} h_2=\frac{v^2}{2g} \\ {h}_2=\frac{{900}^2}{2×9.8}m \\ {h}_2=41326.53m \end{gathered}$$

Thus, the maximum altitude reached by the rocket is,

$$\begin{gathered} H=h_1+h_2 \\ H=\left(13500+41326.53\right)m \\ H=54,826.53m \\ H=54.83km \end{gathered}$$

In the first phase, the rocket accelerates for $t=30s$.

In the second phase, the rocket moves at a constant speed of $900m/s$after running out of the fuel.

Maximum altitude covered by the rocket is,$H=54,826.53m$

In the second phase, the rocket moves with a speed of $v=900m/s.$

So the time consumed during this phase is,

$$\begin{gathered} t=\frac{v}{g} \\ t=\frac{900}{9.8}s \\ t=91.84s \end{gathered}$$

Also, the time taken to cover the maximum altitude can be determined by, role="math" localid="1648537958131" $t_2=\sqrt{\frac{2H}{g}}$

Therefore,

$$\begin{gathered} t_2=\sqrt{\frac{2×54,826.53}{9.8}}s \\ {t}_2=105.78s \end{gathered}$$

Hence, the rocket elapsed a total time of $T=t+t_1+t_2$in the air.

Therefore,

$$\begin{gathered} T=\left(30+91.84+105.78\right)s \\ T=227.62s \end{gathered}$$