Question

2) A hot air balloon rising vertically is being tracked by an observer on the ground, located 60 meters from the take-off point. At a certain instant, the balloon is 30 meters above the ground and the angle of elevation (between the ground and the observer’s sight line) is increasing at the rate of 0. 1 rad./sec. How fast is the balloon rising at that instant?

Answer 1

let the height of the balloon be h m
then tanØ = h/60 , where Ø is the angle of elevation
h = 60 tanØ
dh/dt = sec^2 Ø dØ/dt

when h = 30, dØ/dt = .1 rad/sec
then tanØ = 30/60 = 1/2
hypotenuse = 30√5
secØ = 30√5/60 = √5/2
sec^2 Ø = 5/4

dh/dt = (5/4)(.1) m/s
= .125 m/sec or 12.5 cm/second

check my arithmetic