For the amusement of the guests, some hotels have elevators onthe outside of the building. One such hotel is 300 feet high. Youare standing by a window 100 feet above the ground and 150 feetaway from the hotel, and the elevator descends at a constant speedof 30 ft/sec, starting at time
t = 0,
where t is time in seconds. Let θ be the anglebetween the line of your horizon and your line of sight to theelevator.
(a) Find a formula for
h(t),
the elevator's height above the ground as it descends from thetop of the hotel.
h(t) =Â Â
 Â
(b) Using your answer to part (a), express θ as a functionof time
t.
θ(t) = Â
tan−1(2−t5​)
Find the rate of change of θ with respect tot.
=Â Â
(c) The rate of change of θ is a measure of how fast theelevator appears to you to be moving. At what height is theelevator when it appears to be moving fastest?
h =Â Â
