Science

To estimate the density of air $\rho$, we can rearrange the terminal velocity formula:

$$|v_{\text{terminal}}| = \sqrt{\frac{2mg}{C_D \rho A}}$$

We can solve this equation for $\rho$:

$$\rho = \frac{2mg}{C_D A |v_{\text{terminal}}|^2}$$

Given that $g \approx 9.8 \, \text{m/s}^2$ and $C_D \approx 1$, we can use our measurements for $m$, $A$, and $|v_{\text{terminal}}|$ to estimate $\rho$.

$$\rho = \frac{2 \times 0.1 \times 9.8}{1 \times 0.01 \times 5^2}$$

$$\rho \approx \frac{1.96}{0.25}$$

$$\rho \approx 7.84 \, \text{kg/m}^3$$

So, based on these measurements and estimates, the density of air $\rho$ is approximately $7.84 \, \text{kg/m}^3$.