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
Originally Posted by BarlowBarlow1
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
the question as asked.
Now the rock appears to be moving most rapidly when da/dt is a maximum,
which will be a point at which d^2a/dt^2 = 0.