pre-cal

if the helicopter is moving forward at 30 mph and drops 400 feet in the elevation while the vehicle is moving 48 mph, how does the angle of depression change after 10 minutes?

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  1. So the helicopter drops 400 feet. How high was it when it started its descent?

    what is "the vehicle"? Is it a car driving directly underneath the chopper?

    Did we start at t=0 when the chopper was directly above "the vehicle"?

    Which angle of depression? chopper-to-ground? chopper-to-fixed point on ground? chopper-to-moving "vehicle"?

    Woefully incomplete conditions!

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    posted by Steve
  2. However, assuming the chopper starts at (0,h) and the vehicle starts at (0,0),

    after t minutes the chopper and vehicle are at

    (2640t,h - 5/33 t) and
    (4224t,0)

    So, the line between them has slope

    (h-5/33 t)/(1584t)

    Now you have the angle

    tanθ = (h-5/33 t)/(1584t)
    so you can figure dθ/dt at t=10.

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    posted by Steve

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