a spy uses a telescope to track a rocket launched vertically from a launching pad 6 kmaway. At a certain moment, the angle (θ) between the telescope and the the ground is equal to π3 and is changing at a rate of 0.9 rad/min. What is the rocket's velocity atthat moment?

v = 21.6 Km/min

Step-by-step explanation:

x = 6 Km

∅ = π/3 (variable)

d∅/dt = 0.9 rad/min

Let ∅ be the angle between the spy's line of sight and the ground and y is the distance between the rocket and the ground.

Then we have the equation

tan ∅ = y / x

⇒ (tan ∅) ' = (y / x) ' = (1/x) * y'

⇒ Sec²∅ * (d∅/dt) = (1/x) * (dy/dt) = (1/x) * v

⇒ v = x*Sec²∅ * (d∅/dt)

⇒ v = 6 Km*Sec² (π/3) * (0.9 rad/min)

⇒ v = 21.6 Km/min