# Related Rate

• May 20th 2007, 09:52 AM
BarlowBarlow1
Related Rate
A mountain climber on one edge of a deep canyon 800 ft wide sees a large rock fall from the opposite edge at time=0. As he watches the rock plummet downward, his eyes first move slowly, then faster, and than more slowly again. Let a be the angle of depression of his line of sight below the horizontal. At what angle a would seem to be moving the most rapidly. That is, would da/dt be maximal? Remember the rocks acceleration would be -32 ft/sec^2
• May 20th 2007, 11:00 AM
CaptainBlack
Quote:

Originally Posted by BarlowBarlow1
A mountain climber on one edge of a deep canyon 800 ft wide sees a large rock fall from the opposite edge at time=0. As he watches the rock plummet downward, his eyes first move slowly, then faster, and than more slowly again. Let a be the angle of depression of his line of sight below the horizontal. At what angle a would seem to be moving the most rapidly. That is, would da/dt be maximal? Remember the rocks acceleration would be -32 ft/sec^2

Take the climber's height be takes as the height reference, that is we take
a coordinate system with the climber at h=0.

Let the initial height of the rock he h1, then as it falls its height is:

s=-16 t^2 + h1.

We also have the angle of depression a satisfies the equation:

tan(a) = s/800 = [16 t^2 + h1]/800.

Now you may be expected to assume that h1=0, but I don't see that in