offee filters behave Ince small parachutes, with a drag force that is proportional to the velocity squared, \(F_{\text {drag }}=K v^{2}\). A single coffee filter, when dropped from a height of \(2.0 \mathrm{~m}\), reaches the ground in a time of \(3.0 \mathrm{~s}\). When a second coffee filter is nestled within the first, the drag force remains the same, but the weight is doubled. Find the time for the combined filters to reach the ground. (Neglect the brief period when the filters are accelerating up to their terminal speed.)

We start by analyzing the motion of a single coffee filter. It falls from a height of 2.0 m and takes 3.0 s to reach the ground. We are given that \(F_{\text {drag }}=K v^{2}\), where K is the constant of proportionality. First, we will find the terminal velocity for a single coffee filter. As the motion is almost a free fall with a drag force, we can write Newton's second law in the vertical direction as \(F_{\text {drag }}=m \cdot g\), where m is the mass and g is the acceleration due to gravity. Since the motion is at terminal velocity, \(F_{\text {drag }}\) is balanced by the weight of the coffee filter. Hence, \(K v^{2}=m \cdot g\) or \(v^{2} = \frac{m \cdot g}{K}\). We do not know K yet, but we will find it in step 2.

Now, we need to find the drag force constant K. We are given the time it takes for a single coffee filter to drop from a height of 2.0 m which is 3.0 seconds. So we can use the equation, \(y = vt\), where y is the distance fallen, v is the terminal velocity, and t is the time it takes to reach the ground. We will substitute v from step 1 and solve for K. \(2.0 \, \text{m} = \sqrt{\frac{m \cdot g}{K}} \cdot 3.0 \,\text{s}\) Solving for K, \(\displaystyle K = \frac{m \cdot g}{ \left(\frac{2.0 \mathrm{~m}}{3.0 \mathrm{~s}}\right)^2}.\)

Next, we will analyze the motion of the combined filters that are nestled within each other. The total weight of the combined filters is doubled, so the equation becomes \(F_{\text {drag }}=2m \cdot g\). The drag force remains the same, so we can write \(K v'^{2}=2m \cdot g\), where \(v'\) is the new terminal velocity. Dividing this equation by the equation from step 1, we get \(\displaystyle \frac{v'^{2}}{v^{2}} = 2\). Therefore, the terminal velocity of the combined filters, \(v' = v \sqrt{2}\).

Finally, we need to find the time it takes for the combined filters to reach the ground. We can use the equation \(y = v't'\), where y is the distance fallen, and \(t'\) is the time it takes to reach the ground. \(2.0 \, \text{m} = \sqrt{2} \cdot v \cdot t'\) We will substitute v from step 1, and solve for \(t'\). \(\displaystyle t' = \frac{2.0 \mathrm{~m}}{\sqrt{2} \cdot \sqrt{\frac{m \cdot g}{K}}} = \frac{3.0 \, \text{s}}{\sqrt{2}} \approx 2.12 \,\text{s}\). So, the time for the combined filters to reach the ground is about 2.12 seconds.