an observer stands 200 m from the launch site of a hot-air balloon. the balloon rises vertically at a constant rate of 4 m/s. how fast is the angle of elevation increasing 30 s after the launch.

Answer 1

tan(a) = h/200

sec^2(a) da/dt = 1/200 dh/dt

da/dt = 1/200 cos^2(a) dh/dt

When t=30, h = 4*30 = 120m
tan(a) = 120/200 = 0.6
so, cos(a) = 0.857

da/dt = .005 * 0.857^2 * 4 = .0147 rad/s

That's roughly pi/68 rad/s

Answer 2

Were you drunk when you posted this Steve?