Arocket is launched vertically from a point on the ground. an observer who is 1000 m away from the base of the launching pad, notices that the angle of elevation of the rocket is increasing at a rate of π/40 radians per second when the angle of elevation is π/3.
find the speed of the rocket at that instant.

dh/dt=18.75rad/sec

Explanation:

A rocket is launched vertically from a point on the ground. An observer who is 1000 m away from the base of the launching pad, notices that the angle of elevation of the rocket is increasing at a rate of π/40 radians per second when the angle of elevation is π/3.

Find the speed of the rocket at that instant.

you should have a right-angled triangle, where the angle of elevation is Ø, the adjacent is 1000 and the opposite is h, the height of the rocket.

tangent of an angle is opposite over adjacent side

tanØ = h/100 0

h = 1000tanØ

differentiate both sides with respect to dt,change in time

tanØ=sec^2 Ø

dh/dt = 1000sec^2 Ø (dØ/dt)

where Ø = 60° or π/3 rad

dØ/dt = π/40 radians per second

substituting the value into the equation

dh/dt=1000sin^260° *π/40 rad/sec

dh/dt=750*π/40 rad/sec

dh/dt=18.75rad/sec

dh/dt is the same as speed, change in height per time

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr