Arocket is launched at the rate of 11 feet per second from a point on the ground 15 feet from an observer. to 2 decimal places in radians per second, find the rate of change of the angle of elevation when the rocket is 30 feet above the ground. type your answer in the space below. if your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Supposing the person has insignificant height, we have that the angle of elevation is connected to the height of rocket and distance from spectator by:
tan(θ) = H/D (opposite is height, adjacent is distance)
Taking the derivative of both sides with respect to time:
d/dt (tan(θ(t))) = v/D (where v is the velocity of the firework)
Using chain rule on the left side:
sec2(θ)*(dθ/dt) = v/D
dθ/dt = cos2(θ)*v/D
When the firework is 30 feet above the ground, θ = tan-1(30ft/15ft) = 0.540 rad
dθ/dt = [ 1 / (1 + 1) ]*(11)/15 
= (1/2)(1/15)*(11) 
= 11/100 
= 0.011 radians /s

The low lands i believe