Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus & Beyond Homework (http://www.physicsforums.com/forumdisplay.php?f=156)
-   -   Rate of change of height and angle of elevation. (http://www.physicsforums.com/showthread.php?t=260738)

 bri.nguy Oct1-08 12:14 AM

Rate of change of height and angle of elevation.

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)
h'=200sec^2(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.

 tiny-tim Oct1-08 04:43 AM

Welcome to PF!

Quote:
 Quote by bri.nguy (Post 1895912) Rate of change of height wrt angle of elevation h=200tan (thetha) h'=200sec^2(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.
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')

 All times are GMT -5. The time now is 10:39 PM.

Powered by vBulletin Copyright ©2000 - 2013, Jelsoft Enterprises Ltd.
© 2012 Physics Forums