A boy on level ground watches a hot air balloon launched from a point 18.0 m away from him. The balloon rises with constant speed 0.314 m/s. Find the rate at which the angle of elevation of the balloon at his eye is changing when the angle becomes 60.0 degrees.
Answers
Given : A boy on level ground watches a hot air balloon launched from a point 18.0 m away from him. The balloon rises with constant speed 0.314 m/s.
To Find : the rate at which the angle of elevation of the balloon at his eye is changing when the angle becomes 60.0 degrees.
Solution:
α = angle of elevation
tan α = Height / Horizontal distance
tan α = x / a
x = Height
a= Horizontal distance = 18 m
tan α = x / a
sec²α dα/dt = (dx/dt )/a
dα/dt = (dx/dt )/a sec²α
dx/dt = 0.314 m/s
a =18
sec α = sec 60° = 2 => sec²α = 4
=> dα/dt = 0.314 / ( 18 * 4)
=> dα/dt = 0.004361
rate at which the angle of elevation of the balloon at his eye is changing = 0.004361
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