91Ó°ÊÓ

• Mass of spherical ball = 6.00 kg.
• Radius of the ball = R = 3.00 cm = 0.03 m.
• Drag coefficient = C = 1.60.
• The density of air medium $\mathrm{ÒÏ}=1.20\mathrm{kg}/{\mathrm{m}}^3$.

When a blunt body, like a spherical ball, falls from rest through the air, the drag force is upward, and its magnitude increases as the speed of the body increase until the drag force eventually equals (or balances) the downward gravitational force. Then the body moves with a constant speed called terminal speed which is given by:

$v_t=\sqrt{\frac{2F_g}{C\mathrm{ÒÏ}A}}$

Where

$$\begin{gathered} C_P=\mathrm{experimental}\mathrm{determined}\mathrm{drag}\mathrm{coefficient};A=\mathrm{cross}\mathrm{sectional}\mathrm{area}\mathrm{of}\mathrm{body} \\ \mathrm{ÒÏ}=\mathrm{destiny}\mathrm{of}\mathrm{the}\mathrm{air}\mathrm{medium} \end{gathered}$$

$$\begin{gathered} v_t=\sqrt{\frac{2F_g}{C\mathrm{ÒÏ}A}} \\ {v}_t=\sqrt{\frac{2mg}{C\mathrm{ÒÏ}\mathrm{Ï€}{\mathrm{R}}^2}} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v_t=\sqrt{\frac{2×\left(6.00\mathrm{kg}\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{1.60×1.20\mathrm{kg}/{\mathrm{m}}^3×3.14×{\left(0.03\mathrm{m}\right)}^2}} \\ {v}_t=\sqrt{\frac{117.6\mathrm{N}}{5.43×{10}^{-3}\mathrm{kg}/\mathrm{m}}} \\ {v}_t=147\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the terminal speed is 147 m/s.