A weather balloon is launched from a point $d = 453$ m from the observer. It rises vertically at a rate of $v = 1.8$ m/s. The angle of elevation $\theta$ is measured with respect to the horizontal as shown in the figure. a. When the height is $h = 309$ m, the angle $\theta$ is b. At height $h = 309$ m the rate of change in $\theta$ is
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Differentiating both sides with respect to time $t$, we get $\sec^2 \theta \frac{d\theta}{dt} = \frac{1}{d} \frac{dh}{dt}$ $\frac{d\theta}{dt} = \frac{\cos^2 \theta}{d} \frac{dh}{dt}$ Given that $d = 453$ m and $\frac{dh}{dt} = v = 1.8$ m/s. When $h = 309$ m, we Show more…
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