Answer:
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