A free-falling skydiver of mass 80 kg experiences viscous friction proportional to velocity, where the constant of proportionality is 0.3 . Taking gravity as \( 10 \mathrm{~m}^{2} / \mathrm{s} \), and letting \( v \) represent the velocity of the skydiver, the dynamic model for a skydiver in freefall is given by: \( \begin{array}{llll}\text { (A) } \frac{d v}{d t}+0.3 v^{2}=80 & \text { (B) } 80 \frac{d v}{d t}+0.3 v^{2}=800 & \text { (C) } 80 \frac{d v}{d t}+v^{2}=240 \\ \text { (D) } 8 \frac{d v}{d t}+0.3 v=80 & \text { (E) } 80 \frac{d v}{d t}+0.3 v=800 & \text { (F) } 80 \frac{d v}{d t}+0.3 v=240\end{array} \)

To understand this scenario better, let's visualize our skydiver gracefully plummeting through the air. In this case, the forces at play include the force of gravity, which is pulling the skydiver downward with a force of \(80 \times 10 = 800 \, \text{N}\), and the viscous friction opposing this motion, which is proportional to the velocity (given as \(0.3v\)). The model should balance the force of gravity and the drag force. It's key to note that the rate of change of velocity (\(\frac{d v}{d t}\)) is directly related to the net force acting on our skydiver. Hence, the correct formulation involves both the mass of the skydiver and the viscous drag combined, leading us to option (E): \(80 \frac{d v}{d t} + 0.3 v = 800\). Keep that in mind for future free-fall moments!