P01R2015 – Rising Rocket

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 $\left(v_{0y}=0\right)$, the kinematic equations of the rocket are

\left\{ \begin{array}{c}
\displaystyle{y\left(t\right)=\frac{1}{2}a_yt^2+v_{0y}t+y_0=\frac{1}{2}a_yt^2} \\
\\
v_y\left(t\right)=a_yt+v_{0y}=a_yt \end{array}
\right.

Note: the acceleration $a_y$ given is the net upward acceleration of the rocket, you do not need to subtract gravity $g$ from it.

We then go over two different methods to derive the speed of the rocket when it runs out of fuel.

Method 1: solving for time and plugging into $v_y\left(t\right)$

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 derive that the rocket reaches a height $h$ at a time $t_1$ equal to

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

The speed of the rocket at that time is equal to

\boxed{v_y\left(t_1\right)=a_yt_1=a_y\sqrt{\frac{2h}{a_y}}=\sqrt{2a_yh}}

Method 2: solving for $v_y$ directly
The rocket runs out of fuel at height $h$ and therefore its speed at that instant can be derive directly as follows

\begin{aligned}
v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ &\Rightarrow \ \ \ \ \ v^2_y=0+2a_yh \\
&\Rightarrow \ \ \ \ \ v_y=\pm \sqrt{2a_yh}
\end{aligned}

Keeping only the positive root (because speed is positive), we conclude that the speed of the rocket when it runs out of fuel is equal to

\boxed{v_y=\sqrt{2a_yh}}

Based on the previous question, we can say that the rocket runs out of fuel at time $t_1$ equal to

\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 Therefore, the kinematic equations established previously are no longer valid and we setup new equations for this phase of the motion.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.

We then go over two different methods to derive the maximum height reached by the rocket.

Method 1: solving for time and substituting into $y\left(t\right)$
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 maximum height reached by the rocket is given by $H_{max}=h+H$ and is equal to

\boxed{H_{max}=h+H=\left(1+\frac{a_y}{g}\right)h}

Method 2: solving for $H$ directly
The rocket reaches its maximum height when its velocity cancels (turn-around point) which requires that it travel an additional height $H$ that satisfies the following

\begin{aligned}
v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ &\Rightarrow \ \ \ \ \ 0={\left(\sqrt{2a_yh}\right)}^2+2\left(-g\right)\left(H\right) \\
\\
&\Rightarrow \ \ \ \ \ 0=2a_yh-2gH \\
\\
&\Rightarrow \ \ \ \ H=\frac{a_yh}{g}
\end{aligned}

The maximum height reached by the rocket is given by $H_{max}=h+H$ and is equal to

\boxed{H_{max}=h+H=\left(1+\frac{a_y}{g}\right)h}