Answer:
[tex]-0.288\ \text{rad/s}[/tex]
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
[tex]\dfrac{dy}{dt}[/tex] = Rate of change in height of rocket = 12 m/s
[tex]\dfrac{dx}{dt}[/tex] = Rate of change in x = 0
Distance between observer and rocket at y = 200 m
[tex]r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{150^2+200^2}\\\Rightarrow r=250\ \text{m}[/tex]
[tex]\tan\theta=\dfrac{y}{x}[/tex]
Differentiating with respect to time
[tex]\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}}{\sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{0-150\times 12}{150^2}}{(\dfrac{250}{150})^2}\\\Rightarrow \dfrac{d\theta}{dt}=-0.288\ \text{rad/s}[/tex]
The rate of change of the angle of elevation is [tex]-0.288\ \text{rad/s}[/tex].
