Rate of change of height and angle of elevation.

  1. 1. The problem statement, all variables and given/known data

    A balloon is rising vertically from a point on the ground that is 200m from an observer at ground level. The observer determines that the angle of elevation between the observer and the balloon is increasing at a rate of 0.9 degrees/s when the angle of elevation is 45 degrees. How fast is the balloon rising at this time?

    2. Relevant equations

    Don't know, other than trig ratios and derivatives.

    3. The attempt at a solution

    Rate of change of height wrt angle of elevation

    h=200tan (thetha)

    (thetha) = 45

    h' = 400 m/degree

    400m/degree * .9 degree / s = 360 m/s

    Now intuitively this makes no sense because, the difference in height of 45 and 45.9 degrees is about 6m not remotely close to 400....Please help, thanks in advance.
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi bri.nguy! Welcome to PF! :smile:

    (have a theta: θ :smile:)

    You need to convert θ from degrees into radians. :wink:

    (and btw, it should be h' = 200sec^2θ θ')
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