An observer stands 150 meters from a fireworks display rocket, which is fired directly upward. When the rocket reaches a height of 200 meters, it is traveling at a speed of 12 meters/second. At what rate is the angle of elevation formed with the observer increasing at that instant

Explanation:

x = Distance of observer from the initial location of the rocket = 150 m

y = Vertical displacement of the rocket from the ground = 200 m

r = Distance between observer and rocket

= Rate of change in height of rocket = 12 m/s

= Rate of change in x = 0

Distance between observer and rocket at y = 200 m

Differentiating with respect to time

The rate of change of the angle of elevation is .

the answer is 46 mls because when you add the speed per second it increses the speed

d is the correct answer