A rocket travels vertically at a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located 16km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after lift-off. Round your answer to the nearest hundredth rad/hr

w = 0.01 t⁻²   rad/h

Explanation:

Let's use trigonometry to find the elevation angle of the telescope, we assume that the rocket remains fixed in the telescope.

        tan θ = CO / CA

the adjacent leg is CA = 16 km = 16 10³ m

Let's use kinematics to find the height of the rocket

          v = y / t

          y = v t

this height is equal to the opposite leg

          CO = v t

we substitute

         tan θ = vt / 16

         tan θ = 1200/16 t

         tan θ = 75 t

         θ = tan⁻¹ ( 75 t )

speed is defined by

          w = dθ/dt

          w =  

as time increases we can neglect the 1 of the denominator

          w = 1/75 t²

          w = 0.0133 t⁻²

Using three significant figures

           w = 0.01 t⁻²   rad/h

for this answer the time must be given in hours

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow