Question

An airplane at a height of 2000 meters is flying horizontally, directly toward an observer on the ground, with a speed of 300 meters per second. How fast is the angle of elevation of the plane changing when this angle is 45 degrees?

Answer 1

At 45 degree elevation angle, airplane altitude h equals horizontal distance from plane to observer, d. (d is the distance from the plane to a point directly above the observer).

A = tan-1 (h/d) is the elevation angle.

dA/dt = d/dh [tan-1 (h/d)]* dh/dt
= V* d/dh [tan-1 (h/d)]
= V*[1/(1 + (h/d)^2)]*(1/d)

When h = d = 2000 m,

dA/dt = (300 m/s)*(1/2)*(1/2000 m)
= 0.075 radians/s