Question

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

Answer 1

Answer:

$-0.288\ \text{rad/s}$

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

$\dfrac{dy}{dt}$ = Rate of change in height of rocket = 12 m/s

$\dfrac{dx}{dt}$ = Rate of change in x = 0

Distance between observer and rocket at y = 200 m

$r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{150^2+200^2}\\\Rightarrow r=250\ \text{m}$

$\tan\theta=\dfrac{y}{x}$

Differentiating with respect to time

$\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}}{\sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{0-150\times 12}{150^2}}{(\dfrac{250}{150})^2}\\\Rightarrow \dfrac{d\theta}{dt}=-0.288\ \text{rad/s}$

The rate of change of the angle of elevation is $-0.288\ \text{rad/s}$.