Aspy 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

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17