A rocket is fired vertically into the air at the rate of 6 miles/min. An observer on the ground is located 4 miles from the launching pad. When the rocket is 3 miles high, how fast is the angle of elevation between the rocket and the observer changing? Specify units.

Answer 1

at the moment in question, the rocket is 5 miles from the observer.

h/4 = tanθ
dh/dt = 4 sec^2θ dθ/dt
6 = 4(5/4)^2 dθ/dt
24/25 = dθ/dt

I expect you can deduce the units.

Answer 2

yes steve right