Rate of descent of a hot air-balloon

Question

A rope is attached to the bottom of a hot air-balloon that is floating above a flat field. If the angle of the rope to the ground remains at 65 degrees and the rope is pulled in at 5 ft/s, how quickly is the elevation of the ballon changing?

Answer 1

Since the angle is constantly 65 degrees, you can just use trigonometry to solve for the vertical difference $x$: $$\sin(65^{\circ}) = \frac{5}{x} \Rightarrow x = 5\sin(65^{\circ}) \approx 4.53 ft/s$$

Answer 2

Let length of rope between ground and balloon be $r$. We are given that $\frac{dr}{dt} = -5 \mbox{ ft/s}$. The elevation of the balloon is $Cr$, where constant $C = \sin(65^{\circ}).$

The elevation of balloon is changing at rate $$ \begin{align} \frac{d(Cr)}{dt} & = C \frac{dr}{dt} \mbox{ since $C$ is constant} \\ & = -5C \mbox{ ft/s}. \end{align} $$