Question

Sasha is in the park with her friend megan. They are separated by 50 ft. megan lets go of his red ballon and it goes straight up in the air. At what rate is sasha's eye angle changing if he is watching the ballon rise at a rate of 4 ft per sec, the moment the balloon is 120 ft in the air

Answer 1

if the balloon is at height h,

tanθ = h/50
sec^2θ dθ/dt = 1/50 dh/dt
at h=120,

(5/13)^2 dθ/dt = 4/50
dθ/dt = 4/50 * 169/25 = 676/1250 = 0.54 rad/s