Checking Work from 2-71 in University Physics (Warning: there is a lot of it)

[Ackbach]

This is from Problem 2-71 in Young and Freedman's 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
$$y_{3}=553-9.8(9.12711)^{2}/2=144.8097\,\text{m}.$$

 

[I like Serena]

Just a couple of observations about the initial setup.

When the rocketeer catches the student, they will have different speeds.
For the rocketeer to overtake the student, he will be going much faster when they catch up.
How is that speed difference bridged?

To allow a maximum acceleration of 5g on their bodies, the effective acceleration will have to be 4g.

 

[Ackbach]

Just a couple of observations about the initial setup.

When the rocketeer catches the student, they will have different speeds.
For the rocketeer to overtake the student, he will be going much faster when they catch up.
How is that speed difference bridged?

Their speed functions are different before the Rocketeer catches the student. This is a kinematics problem, not a dynamics problem, so the exact mechanics of how the Rocketeer catches the student, and the resulting sudden change in velocity, is probably not what the problem is interested in.

To allow a maximum acceleration of 5g on their bodies, the effective acceleration will have to be 4g.

I'm not sure I agree. I think the problem is talking about net acceleration, caused by net force. I don't think the acceleration due to the jet pack alone is what they're talking about. Maybe I am incorrect there, but that's the way I read it.

 

[MarkFL]

I have used a different approach, but it disagrees with your result.

I began with the kinematic relationship:

\(\displaystyle a=\frac{v_i^2-v_f^2}{2\left(y_i-y_f \right)}\)

and applied it to the period of time the student and the Rocketeer are slowing down. We are told $a=5g,\,v_f=0,\,y_f=0$ and so we have:

\(\displaystyle 5g=\frac{v_i^2}{2y_i}\)

Now, we want to solve for $y_i$, the height at which the Rocketeer catches the student.

\(\displaystyle 10y_i=\frac{v_i^2}{g}\)

Now, we also know that \(\displaystyle v_i=gt\) and so we have:

\(\displaystyle 10y_i=gt^2\)

And we also know that \(\displaystyle y_i=553-\frac{gt^2}{2}\,\therefore\,gt^2=1106-2y_i\) which gives us:

\(\displaystyle 10y_i=1106-2y_i\)

\(\displaystyle y_i=\frac{1106}{12}=\frac{553}{6}\)

I had doubts about this solution, thinking I must surely be oversimplifying, so I googled the problem and found it agrees with the solution to problem number 2 here:

http://modeling.asu.edu/modeling/DesbienDwain_graphsTPT08.pdf

 

[Jameson]

Very nice answer Mark!

I have been reading and rereading it because it seems too simple, but it looks perfect and after reading the PDF file you linked to I see why. It's interesting that with the model you used you don't have to worry about the initial velocity of the Rocketeer because having a max of $a=5g$ restricts how far from the ground the Student can be. In fact you can think of this problem as if there is only one person. It becomes asking the height at which accelerating upwards at $5g$ from that height will stop the Student perfectly on the ground if starting at 553m.

If you put restrictions on the Rocketeer's motion then it might get more complex. Good problem. :)

 

[I like Serena]

Their speed functions are different before the Rocketeer catches the student. This is a kinematics problem, not a dynamics problem, so the exact mechanics of how the Rocketeer catches the student, and the resulting sudden change in velocity, is probably not what the problem is interested in.

I consider this a serious flaw in Young and Freedman.
Any course in physics should develop the sense of what is physically possible and important in a student.

Anyway, the problem has multiple solutions depending on how this is interpreted.
Does the student suddenly have the same speed as the rocketeer?
Or does the rocketeer suddenly have the speed of the student?
Or do both continue on their own speed, just with a different acceleration (meaning both cannot land at speed zero).

I'm not sure I agree. I think the problem is talking about net acceleration, caused by net force. I don't think the acceleration due to the jet pack alone is what they're talking about. Maybe I am incorrect there, but that's the way I read it.

When a pilot gets G forces on his body, what's important, physically speaking, is the net acceleration on his body.
But I agree that the problem statement appears to be a bit ambiguous.
Another flaw in the book in my opinion.

 

[zzephod]

This is from Problem 2-71 in Young and Freedman's University Physics, 9th Ed.

A determined student walks off the top of ...

It seems to me that there is a problem here in that when the rockateer catches the student a conservation of linear momentum consideration applies to the joint speed of the rockateer and student after capture. For this we need the relative masses of he student and rockateer which we don't have. We could assume that they have equal mass but that seems implausible. However whatever we assume for their masses neither the rockateer's or the student's speeds are continuous at \(\displaystyle t_3\) unless they are already at the same speed.

.

 

[Ackbach]

I have used a different approach, but it disagrees with your result.

I began with the kinematic relationship:

\(\displaystyle a=\frac{v_i^2-v_f^2}{2\left(y_i-y_f \right)}\)

and applied it to the period of time the student and the Rocketeer are slowing down. We are told $a=5g,\,v_f=0,\,y_f=0$ and so we have:

\(\displaystyle 5g=\frac{v_i^2}{2y_i}\)

Now, we want to solve for $y_i$, the height at which the Rocketeer catches the student.

\(\displaystyle 10y_i=\frac{v_i^2}{g}\)

Now, we also know that \(\displaystyle v_i=gt\)

So you're taking the Rocketeer/student combo as having a velocity equal simply to the student's velocity at that point? Read zzephod's comment below: I think there are some unphysical assumptions required here. We would have a completely inelastic collision at the moment of catching. Suppose, as a rough order of magnitude, that the Rocketeer and the student have the same mass. Then the sum of the velocities before the collision must be equal to twice the velocity after the collision.

and so we have:

\(\displaystyle 10y_i=gt^2\)

And we also know that \(\displaystyle y_i=553-\frac{gt^2}{2}\,\therefore\,gt^2=1106-2y_i\) which gives us:

\(\displaystyle 10y_i=1106-2y_i\)

\(\displaystyle y_i=\frac{1106}{12}=\frac{553}{6}\)

I had doubts about this solution, thinking I must surely be oversimplifying, so I googled the problem and found it agrees with the solution to problem number 2 here:

http://modeling.asu.edu/modeling/DesbienDwain_graphsTPT08.pdf

It seems to me that there is a problem here in that when the rockateer catches the student a conservation of linear momentum consideration applies to the joint speed of the rockateer and student after capture. For this we need the relative masses of he student and rockateer which we don't have. We could assume that they have equal mass but that seems implausible. However whatever we assume for their masses neither the rockateer's or the student's speeds are continuous at \(\displaystyle t_3\) unless they are already at the same speed.

.

Exactly. That's what's been troubling me about this whole problem. They can't have the same velocity right before the catching, but they're going to have the same velocity right after. The Rocketeer will be moving faster (perhaps considerably faster) than the student right before the catch. Thank you for so clearly articulating what was only nebulous in my own mind.

 

[MarkFL]

I am assuming that the concept of conservation of linear momentum is introduced some half a dozen or so chapters later in the textbook cited. This notwithstanding, as pointed out, the masses of the student and Rocketeer are not mentioned in any way.

I would easily agree though that during the process of being caught, the velocity of the student must increase, and so the minimal height I found is too small. I think this problem would be more appropriate as a chapter 2 problem if the Rocketeer is said to activate his jet pack prior to catching the student in such a way as to exactly match his speed to that of the student at the moment of the catching. :D

 

[Ackbach]

I am assuming that the concept of conservation of linear momentum is introduced some half a dozen or so chapters later in the textbook cited.

Oddly enough, that's exactly where it is introduced.

This notwithstanding, as pointed out, the masses of the student and Rocketeer are not mentioned in any way.

I would easily agree though that during the process of being caught, the velocity of the student must increase, and so the minimal height I found is too small. I think this problem would be more appropriate as a chapter 2 problem if the Rocketeer is said to activate his jet pack prior to catching the student in such a way as to exactly match his speed to that of the student at the moment of the catching. :D

Right. I'll proceed on that assumption, and mention it in my solution.