A 200 kg weather rocket is loaded with 100 kg of fuel and

fired straight up. It accelerates upward at 30 m/s2 for 30 s, then runs out of fuel. Ignore any air resistance effects.

a. What is the rocket’s maximum altitude?

b. How long is the rocket in the air before hitting the ground?

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,

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

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

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

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,

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

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,

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

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

Therefore,

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