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- Jan 26, 2012

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*University Physics*, 9th Ed.

A determined student walks off the top of the CN Tower in Toronto, which is $553\,\text{m}$ high, and falls freely. His initial velocity is zero. The Rocketeer arrives at the scene five seconds later and dives off the top of the tower to save the student. The Rocketeer leaves the roof with an initial downward velocity of magnitude $v_{0}$ and then is in free fall. In order both to catch the student and to prevent injury to him, the Rocketeer should catch the student at a sufficiently great height above ground so that the Rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by the Rocketeer's jet pack, which he turns on just as he catches the student; before then, the Rocketeer is in free fall. To prevent discomfort to the student, the magnitude of the acceleration of the Rocketeer and the student as they move downward together should be no more than five times $g$. What is the minimum height above the ground at which the Rocketeer should catch the student?

**Answer.**

We must have some careful notation, as there are a number of events we must describe. Let Event $1$ be the student walking off the roof. Let Event $2$ be the Rocketeer pushing himself downward off the roof. Let Event $3$ be the Rocketeer catching the student and igniting his rockets, and let Event $4$ be touchdown on the ground. We will use only one variable for time, $t$. We will use $r$ subscripts for the Rocketeer, and $s$ subscripts for the student. Thus, we have the following

functions that describe the positions, velocities, and accelerations of the student and the Rocketeer:

\begin{align*}

y_{s}(t)&=\begin{cases}y_{1}-gt^{2}/2,\quad &t_{1}<t<t_{3}\\

y_{3}-v_{3}(t-t_{3})+5g(t-t_{3})^{2}/2,\quad &t_{3}<t<t_{4}\end{cases}\\

y_{r}(t)&=\begin{cases}y_{1},\quad &t_{1}<t<t_{2} \\

y_{1}-v_{0}(t-t_{2})-g(t-t_{2})^{2}/2,\quad &t_{2}<t<t_{3}\\

y_{3}-v_{3}(t-t_{3})+5g(t-t_{3})^{2}/2,\quad &t_{3}<t<t_{4}

\end{cases}\\

v_{s}(t)&=\begin{cases}-gt,\quad &t_{1}<t<t_{3}\\

-v_{3}+5g(t-t_{3}),\quad &t_{3}<t<t_{4}\end{cases}\\

v_{r}(t)&=\begin{cases}0,\quad &t_{1}<t<t_{2} \\

-v_{0}-g(t-t_{2}),\quad &t_{2}<t<t_{3}\\

-v_{3}+5g(t-t_{3}),\quad &t_{3}<t<t_{4}

\end{cases}\\

a_{s}(t)&=\begin{cases}-g,\quad &t_{1}<t<t_{3}\\

5g,\quad &t_{3}<t<t_{4}\end{cases}\\

a_{r}(t)&=\begin{cases}0,\quad &t_{1}<t<t_{2} \\

-g,\quad &t_{2}<t<t_{3}\\

5g,\quad &t_{3}<t<t_{4}

\end{cases}.

\end{align*}

The target variable is $y_{3}$. We will set $t_{1}=0$. Then $t_{2}=5\,\text{s}$. Hence, the other unknowns are $t_{3},$ and $t_{4}$, as well as $v_{3}$ and $v_{0}$. Thus, we have $5$ unknowns. How many equations can we come up with? Continuity of $y_{s}$ at $t_{3}$ can give us one equation:

$$y_{1}-gt_{3}^{2}/2=y_{3}.$$

The same can give us an equation for $y_{r}$:

$$y_{1}-v_{0}(t_{3}-t_{2})-g(t_{3}-t_{2})^{2}/2=y_{3}.$$

We also know that $v_{s}(t_{4})=v_{r}(t_{4})=0$. That yields the equation

$$0=-v_{3}+5g(t_{4}-t_{3}).$$

We can also say that the Rocketeer's velocity is continuous at $t_{3}$. Hence, we have that

$$-v_{0}-g(t_{3}-t_{2})=-v_{3}.$$

Moreover, we have that $y_{s}(t_{4})=y_{r}(t_{4})=0$, which yields the equation

$$y_{3}-v_{3}(t_{4}-t_{3})+5g(t_{4}-t_{3})^{2}/2=0.$$

Thus, we have five equations for five unknowns:

\begin{align*}

y_{3}&=y_{1}-gt_{3}^{2}/2\\

y_{3}&=y_{1}-v_{0}(t_{3}-t_{2})-g(t_{3}-t_{2})^{2}/2\\

0&=-v_{3}+5g(t_{4}-t_{3})\\

-v_{3}&=-v_{0}-g(t_{3}-t_{2})\\

0&=y_{3}-v_{3}(t_{4}-t_{3})+5g(t_{4}-t_{3})^{2}/2.

\end{align*}

We can eliminate $v_{3}$ fairly easily, as $v_{3}=5g(t_{4}-t_{3})$. Hence, our system becomes

\begin{align*}

y_{3}&=y_{1}-gt_{3}^{2}/2\\

y_{3}&=y_{1}-v_{0}(t_{3}-t_{2})-g(t_{3}-t_{2})^{2}/2\\

-5g(t_{4}-t_{3})&=-v_{0}-g(t_{3}-t_{2})\\

0&=y_{3}-5g(t_{4}-t_{3})^{2}/2.

\end{align*}

Next, we can solve for $t_{4}-t_{3}$ to obtain

$$t_{4}-t_{3}=\frac{v_{0}+g(t_{3}-t_{2})}{5g}.$$

Plugging this into our system yields

\begin{align*}

y_{3}&=y_{1}-gt_{3}^{2}/2\\

y_{3}&=y_{1}-v_{0}(t_{3}-t_{2})-g(t_{3}-t_{2})^{2}/2\\

0&=y_{3}-\frac{5g}{2}\left(\frac{v_{0}+g(t_{3}-t_{2})}

{5g}\right)^{2}=y_{3}-\frac{(v_{0}+g(t_{3}-t_{2}))^{2}}{10g}

\end{align*}

Rewriting yields

\begin{align*}

y_{3}&=y_{1}-gt_{3}^{2}/2\\

y_{3}&=y_{1}-v_{0}(t_{3}-t_{2})-g(t_{3}-t_{2})^{2}/2\\

y_{3}&=\frac{(v_{0}+g(t_{3}-t_{2}))^{2}}{10g}.

\end{align*}

We solve the second equation for $v_{0}$ to obtain

$$y_{3}-y_{1}+g(t_{3}-t_{2})^{2}/2=-v_{0}(t_{3}-t_{2}),$$

or

$$y_{1}-y_{3}-g(t_{3}-t_{2})^{2}/2=v_{0}(t_{3}-t_{2}),$$

or

$$v_{0}=\frac{2y_{1}-2y_{3}-g(t_{3}-t_{2})^{2}}

{2(t_{3}-t_{2})}.$$

Plugging this into our system yields

\begin{align*}

y_{3}&=y_{1}-gt_{3}^{2}/2\\

y_{3}&=\frac{\left(\frac{2y_{1}-2y_{3}-g(t_{3}-t_{2})^{2}}

{2(t_{3}-t_{2})}+g(t_{3}-t_{2})\right)^{2}}{10g}.

\end{align*}

Let us work on the second equation a bit:

\begin{align*}

y_{3}&=\frac{\left(\frac{2y_{1}-2y_{3}-g(t_{3}-t_{2})^{2}}

{2(t_{3}-t_{2})}+g(t_{3}-t_{2})\right)^{2}}{10g}\\

&=\frac{\left(\frac{2y_{1}-2y_{3}-g(t_{3}-t_{2})^{2}

+2g(t_{3}-t_{2})^{2}}

{2(t_{3}-t_{2})}\right)^{2}}{10g}\\

&=\left(\frac{2y_{1}-2y_{3}+g(t_{3}-t_{2})^{2}}

{2(t_{3}-t_{2})}\right)^{2}\cdot \frac{1}{10g}\\

&=\frac{[2y_{1}-2y_{3}+g(t_{3}-t_{2})^{2}]^{2}}

{40g(t_{3}-t_{2})^{2}}.

\end{align*}

So our system is then

\begin{align*}

y_{3}&=y_{1}-gt_{3}^{2}/2\\

y_{3}&=\frac{[2y_{1}-2y_{3}+g(t_{3}-t_{2})^{2}]^{2}}

{40g(t_{3}-t_{2})^{2}}.

\end{align*}

I think it will be numerically superior to attempt to solve for $t_{3}$ first. We rewrite the system as

\begin{align*}

2(y_{1}-y_{3})&=gt_{3}^{2}\\

2(y_{1}-y_{3})&=2y_{1}-\frac{[2(y_{1}-y_{3})+g(t_{3}-t_{2})^{2}]^{2}}

{20g(t_{3}-t_{2})^{2}}.

\end{align*}

We substitute the first equation into the second to obtain

$$gt_{3}^{2}=2y_{1}-\frac{[gt_{3}^{2}+g(t_{3}-

t_{2})^{2}]^{2}}{20g(t_{3}-t_{2})^{2}}.$$

Rearranging yields

\begin{align*}

20g^{2}t_{3}^{2}(t_{3}-t_{2})^{2}&=

40gy_{1}(t_{3}-t_{2})^{2}-[gt_{3}^{2}+g(t_{3}-

t_{2})^{2}]^{2}\\

20g^{2}t_{3}^{2}(t_{3}-t_{2})^{2}&=

40gy_{1}(t_{3}-t_{2})^{2}-g^{2}[t_{3}^{4}

+2t_{3}^{2}(t_{3}-t_{2})^{2}+(t_{3}-t_{2})^{4}]\\

20gt_{3}^{2}(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})&=

40y_{1}(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})-g[t_{3}^{4}

+2t_{3}^{2}(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})

+(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})^{2}]\\

20gt_{3}^{4}-40gt_{2}t_{3}^{3}+20gt_{2}^{2}t_{3}^{2}&=

40y_{1}t_{3}^{2}-80y_{1}t_{2}t_{3}+40y_{1}t_{2}^{2}-

g[t_{3}^{4}

+2t_{3}^{4}-4t_{2}t_{3}^{3}

+2t_{2}^{2}t_{3}^{2}

+(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})^{2}].

\end{align*}

Now

\begin{align*}(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})^{2}

&=t_{3}^{4}-2t_{2}t_{3}^{3}+t_{2}^{2}t_{3}^{2}

-2t_{2}t_{3}^{3}+4t_{2}^{2}t_{3}^{2}-2t_{2}^{3}t_{3}

+t_{2}^{2}t_{3}^{2}-2t_{2}^{3}t_{3}+t_{2}^{4}\\

&=t_{3}^{4}-4t_{2}t_{3}^{3}+6t_{2}^{2}t_{3}^{2}

-4t_{2}^{3}t_{3}+t_{2}^{4}.

\end{align*}

Hence,

\begin{align*}

t_{3}^{4}

+2t_{3}^{4}-4t_{2}t_{3}^{3}

+2t_{2}^{2}t_{3}^{2}

+(t_{3}^{2}-2t_{2}t_{3}+t_{2}^{2})^{2}

&=t_{3}^{4}

+2t_{3}^{4}-4t_{2}t_{3}^{3}

+2t_{2}^{2}t_{3}^{2}

+t_{3}^{4}-4t_{2}t_{3}^{3}+6t_{2}^{2}t_{3}^{2}

-4t_{2}^{3}t_{3}+t_{2}^{4}\\

&=4t_{3}^{4}

-8t_{2}t_{3}^{3}

+8t_{2}^{2}t_{3}^{2}

-4t_{2}^{3}t_{3}+t_{2}^{4}.

\end{align*}

Plugging this result back into our larger equation above, we obtain

\begin{align*}

20gt_{3}^{4}-40gt_{2}t_{3}^{3}+20gt_{2}^{2}t_{3}^{2}&=

40y_{1}t_{3}^{2}-80y_{1}t_{2}t_{3}+40y_{1}t_{2}^{2}-

g[4t_{3}^{4}-8t_{2}t_{3}^{3}+8t_{2}^{2}t_{3}^{2}

-4t_{2}^{3}t_{3}+t_{2}^{4}]\\

20gt_{3}^{4}-40gt_{2}t_{3}^{3}+20gt_{2}^{2}t_{3}^{2}

&=

40y_{1}t_{3}^{2}-80y_{1}t_{2}t_{3}+40y_{1}t_{2}^{2}-

4gt_{3}^{4}+8gt_{2}t_{3}^{3}-8gt_{2}^{2}t_{3}^{2}

+4gt_{2}^{3}t_{3}-gt_{2}^{4}.

\end{align*}

Therefore,

\begin{align*}

(20g+4g)t_{3}^{4}+(-40gt_{2}-8gt_{2})t_{3}^{3}

+(20gt_{2}^{2}-40y_{1}+8gt_{2}^{2})t_{3}^{2}

+(80y_{1}t_{2}-4gt_{2}^{3})t_{3}

+(gt_{2}^{4}-40y_{1}t_{2}^{2})&=0\\

24gt_{3}^{4}-48gt_{2}t_{3}^{3}

+(28gt_{2}^{2}-40y_{1})t_{3}^{2}

+(80y_{1}t_{2}-4gt_{2}^{3})t_{3}

+(gt_{2}^{4}-40y_{1}t_{2}^{2})&=0.

\end{align*}

This is a quartic in $t_{3}$. Plugging in $g=9.8\,\text{m/s}^{2}$, $t_{2}=5\,\text{s}$, and $y_{1}=553\,\text{m}$ yields the numerical quartic

$$235.2 t_{3}^{4}-2352t_{3}^{3}-15260t_{3}^{2}+216300t_{3}

-546875=0.$$

Let us use Descartes's Rule of Signs to analyze the roots. We have three sign changes, so the number of positive roots is either $1$ or $3$. If we change the sign of all the

odd-power terms, we get

$$235.2 (-t_{3})^{4}+

2352(-t_{3})^{3}-15260(-t_{3})^{2}-216300(-t_{3})

-546875=0.$$

Thus, there is exactly one negative root. We are not interested in this root. We have two possibilities: 1 negative + 1 positive + 2 complex conjugate roots. Or we have 1 negative + 3 positive roots. My calculator yields the latter. The positive roots are

$$t_{3}=\{4.51955\,\text{s},\;5.90268\,\text{s},\;9.12711\,\text{s}\}.$$

The question is, which solution shall we accept? We started the clock with $t_{1}=0$, and $t_{2}=5$. Since it must be that $t_{3}>t_{2}$, the first solution we rule out. Also, since the second solution leaves a bare $0.9\,\text{s}$ for the Rocketeer to catch up, we think that solution unreasonable as well. Therefore, we go with the

$t=9.12711\,\text{s}$ solution, yielding

$$y_{3}=553-9.8(9.12711)^{2}/2=144.8097\,\text{m}.$$