Question

A fly fisherman in Montana approaches the edge of the Smith River Canyon and notices a "switch back" leading down to the river's edge. This switch back is a walking trail made up of 6 sections on an almost perfectly vertical canyon wall, and a sign reads that each section is a 8 degree descent as measured from the horizon ( in the figure is the same for each section). The fisherman decides to estimate how long it will take him to descend the switch back by dropping a stone into the canyon below, and measuring the stone's time of flight with his wrist watch. If he measures the stone's time of flight to be 7.3 seconds, how long (in seconds) will it take him to descend the switchback at a constant speed of 2 meters per second?

Answer 1

Tini, Im going to simplify the problem a bit to make things easier:
We know that the switchback is almost perfectly vertical, so we can ignore any distance we travel "out of" the paper as we draw the path:

Furthermore, because the angles are the same at each switchback, and our speed is constant as we walk down the path, we can use similar triangles to trade our switchback for one long straight path.

Our switchback has thus become:

Now that we have reduced this into a much simpler picture, let's look at how we will solve our problem:
The question asks, how long will it take to walk down the path while travelling at a constant speed?
Constant speed means we can use the following relationship relating distance travelled (r), speed (v), and time travelled (t):
\[t = \frac{ r }{ v }\]
We know v, so now we need to deduce r:
From our triangle in the above picture, we can use trigonometry to find r:
Since
\[\sin (\Theta) = \frac{ y }{ r }\]
We can rearrange to get
\[r = \frac{ y }{ \sin (\Theta) }\]
Where
\[\Theta = 8 \deg\]
Now we need y:
From dropping the stone from a height y, we measured that it took
\[T = 7.3s\]
We can use kinematics to find y from this:
\[y = \frac{ 1 }{ 2 }gT^2\]
We've just about got it. all that remains is to put these equations together:
\[r = \frac{ y }{ \sin (\Theta) } = \frac{ \frac{ 1 }{ 2 }gT^2 }{ \sin (\Theta) } =\frac{ gT^2 }{ 2*\sin(\Theta) }\]
and \[t = \frac{ r }{ s } = \frac{ \frac{ gT^2 }{ 2*\sin(\Theta) } }{ v }\]
or more simply:
\[t = \frac{ g*T^2 }{ 2*v*\sin(\Theta) }\]
We have now reduced this problem to one that involves only the time the stone fell, g, the speed at which we will walk, and the angle of the switchbacks, all of which are known to us. I hope this helps!

Answer 2

And it looks like a lot of my initial replay got cut off. I'm not sure if the "Draw" function on this site works. Basically, all I did was draw the switchback. From there, I used the fact that the angles are all the same (8 degrees) and used some geometry and similar triangles to convert a messy looking switchback picture into an equivalent-length path that descends with no switchbacks. The triangle has the angle Theta, a hypotenuse going down and left of a length r, and a vertical leg dropping a distance y down to the river. There are some very well drawn stick figures in there, so it is a real shame if you cant see them. Sorry for the inconvenience!