Answer:
0.15 rad/s
Step-by-step explanation:
A rocket is launched at the rate of 11 feet per second from a point on the ground 15 feet from an observer.
To find rate of change of angle when the rocket is 30 feet above the ground
Height of rocket, y = 30 feet
Speed of rocket, [tex]\dfrac{dy}{dt}= 11 ft/s[/tex]
Distance of observer from rocket, d = 15 feet
Angle between observer and rocket at height 30 feet is Ф
[tex]\tan\theta=\dfrac{30}{15}=2[/tex]
[tex]\tan\theta=\dfrac{\text{Perpendicular}}{\text{base}}[/tex]
Distance between observer and rocket launch doesn't change.
So, d=15 will remain constant.
[tex]\tan\theta=\dfrac{y}{d}[/tex]
[tex]15\tan\theta=y[/tex]
differentiate w.r.t t
[tex]15\sec^2\theta \dfrac{d\theta}{dt}=\dfrac{dy}{dt}[/tex]
[tex]15(1+\tan^2\theta)\cdot \dfrac{d\theta}{dt}=11[/tex]
[tex]\dfrac{d\theta}{dt}=\dfrac{11}{15(1+2^2)}[/tex]
[tex]\dfrac{d\theta}{dt}=\dfrac{11}{75}\ rad/s\approx 0.15\ rad/s[/tex]
Hence, The rate of change of the angle of elevation is 0.15 rad/s
