P01G2019 – Rocket Launch

We first set up the kinematic equations of the rocket for the first phase of its motion (until it runs out of fuel).

Considering that the origin is at ground level $\left(y_0=0\right)$ and that rocket is initially at rest $\left(v_0=0\right)$, the kinematic equations of the rocket are:

\left\{ \begin{array}{c}
\displaystyle{y\left(t\right)=\frac{1}{2}a_yt^2+v_0t+y_0=\frac{1}{2}a_yt^2} \\
\\
v_y\left(t\right)=a_yt+v_0=a_yt \end{array}
\right.

The rocket runs out of fuel at altitude $h$ i.e. at a time $t_1$ such that $y\left(t_1\right)=h$. Thus, we solve

\begin{aligned}
y\left(t_1\right)=h\ \ \ \ &\Rightarrow \ \ \ \ \ \ \ \frac{1}{2}a_yt^2_1=h \\
\\
&\Rightarrow \ \ \ \ \ \ \ t_1=\pm \sqrt{\frac{2h}{a_y}}
\end{aligned}

By keeping the positive root only, we have derive that

\boxed{t_1=\sqrt{\frac{2h}{a_y}}}

After $t_1$, the acceleration of the rocket is no longer $a_y$ and instead becomes equal to $-g$ (free-fall). Therefore, the kinematic equations established previously are no longer valid and we setup new equations for this phase of the motion.

Keeping the origin at ground level $\left(y_0=0\right)$ and recalling that the initial speed of the rocket is now $v_0=\sqrt{2a_yh}$, we derive the following kinematic equations

\left\{ \begin{array}{c}
\displaystyle{y\left(t\right)=\frac{1}{2}a_yt^2+v_0t+y_0=-\frac{1}{2}gt^2+\sqrt{2a_yh}\cdot t} \\
\\
v_y\left(t\right)=a_yt+v_0=-gt+\sqrt{2a_yh} \end{array}
\right.

The rocket reaches its maximum height at time $t_2$ (measured from the moment it runs out of fuel) such that $v_y\left(t_2\right)=0$ and we therefore solve

\begin{aligned}
v_y\left(t_2\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -gt_2+\sqrt{2a_yh}=0 \\
\\
&\Rightarrow \ \ \ \ \ t_2=\frac{\sqrt{2a_yh}}{g}
\end{aligned}

In that amount of time, it travels a height $H$ equal to

\begin{aligned}
H=y\left(t_2\right)&=-\frac{1}{2}gt^2_2+\sqrt{2a_yh}\cdot t_2 \\
\\
&=-\frac{1}{2}g\left(\frac{2a_yh}{g^2}\right)+\sqrt{2a_yh}\cdot \frac{\sqrt{2a_yh}}{g} \\
\\
&=-\frac{a_yh}{g}+\frac{2a_yh}{g} \\
\\
&=\frac{a_yh}{g}
\end{aligned}

The total height reached by the rocket is therefore equal to

\boxed{y_{tot}=h+H=\left(1+\frac{a_y}{g}\right)h=\left(1+\frac{3.2}{9.8}\right)\cdot 950\approx 1260\ m}

The rocket falls from its maximum height where $v_y=0$ and reaches a speed $v_{fy}$ just before hitting the ground that is equal to

\begin{aligned}
v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ &\Rightarrow \ \ \ \ \ v_{fy}=\sqrt{2a_y\mathrm{\Delta }y} \\
\\
&\Rightarrow \ \ \ \ \ v_{fy}=\sqrt{2\cdot \left(-g\right)\cdot \left(0-y_{tot}\right)}
\end{aligned}

The speed of the rocket just before it hits the ground is equal to

\boxed{v_{fy}=\sqrt{2gy_{tot}}=\sqrt{2\cdot 9.8\cdot 1260}\approx 157\ m/s}

The rocket falls from its maximum height $H+h$ from rest and therefore its position $y\left(t\right)$ during the fall is given by

y\left(t\right)=-\frac{1}{2}gt^2+H+h

To rise to its maximum height takes the rocket an amount of time $t_1+t_2$. It then falls from its maximum height $H+h$ to the ground from rest which takes an amount of time $t_3$ such that $y\left(t_3\right)=0$.

The time $t_3$ is thus given by

-\frac{1}{2}gt^2_3+H+h=0\ \ \ \ \Rightarrow \ \ \ \ \ t_3=\sqrt{\frac{2\left(H+h\right)}{g}}

and we therefore conclude that the amount of time spent in the air is equal to

\boxed{\mathrm{\Delta }t_{tot}=t_1+t_2+t_3=\sqrt{\frac{2h}{a_y}}+\frac{\sqrt{2a_yh}}{g}+\sqrt{\frac{2\left(H+h\right)}{g}}=\sqrt{\frac{2\cdot 950}{3.2}}+\frac{\sqrt{2\cdot 3.2\cdot 950}}{9.8}+\sqrt{\frac{2\cdot 1260}{9.8}}\approx 48.36\ s}