Answer:
(a) h(t) = 300 - 30t
(b) θ = ((200 - 30t)150)
(c) t =
h = 100m above ground
Step-by-step explanation:
(a)
change of distance of the elevator is governed by equation of speed
speed = distance/time
distance = speed x time
= 30 x t = 30t
since the elevator starts at t = 0 on top of the hotel at 300ft, and the distance between the elevator keep decreasing to the ground, the equation becomes
h(t) = 300 - 30t
(b)
θ = angle between the line of horizon of the observer to the line of sight to elevator
since the observer is 100 feet above the ground, the vertical distance between observer and the top of the hotel is 200ft.
hence, the opposite side of the angle θ is 300 - 30t - 100 = 200 - 30t
therefore, tan θ = (200 - 30t)/150
and θ = ((200 - 30t)150)
Differentiate θ with respect to time, t.
Use chain rule:
let u = (200 - 30t)/150
u' = -1/5
θ = u
θ' =
=
(c)
The elevator will seem to be moving the fastest at θ = 0 since that is the fastest change of θ as the elevator move downward
tan θ = (200 - 30t)/150
tan (0) = 0
(200 - 30t) = 0
200 = 30t
t = 200/30 =
height = 100m above ground