Question

A sphere of diameter D, moving at speed v in air, experiences a quadratic drag force given approximately by FD = (0.25 Ns2/m4) D2 v2, for diameters of few centimetres or more, and speeds of a metre per second or more. Calculate the terminal velocity of a basketball (25 cm diameter, mass 1 kg), and of a grapefruit 10 cm in diameter (assume a density of 1000 kg/m3 for the grapefruit).

Answer 1

Answer on Question #64729, Physics / Mechanics | Relativity

A sphere of diameter $D$, moving at speed $v$ in air, experiences a quadratic drag force given approximately by $F_{D} = (0.25 \, \text{Ns}^{2} / \text{m}^{4}) \, D^{2} \, v^{2}$, for diameters of few centimetres or more, and speeds of a metre per second or more. Calculate the terminal velocity of a basketball (25 cm diameter, mass 1 kg), and of a grapefruit 10 cm in diameter (assume a density of 1000 kg/m³ for the grapefruit).

Solution:

An object falling through the air will reach a terminal velocity when the drag force is equal to the weight:

$$F_{net} = mg - F_{D} = 0$$

So,

$$mg = 0.25 D^{2} v^{2}$$

The terminal velocity from given can be expressed by

$$v_{t} = \sqrt{\frac{mg}{0.25 D^{2}}} = \frac{2}{D} \sqrt{mg}$$

For a basketball

$$v_{t} = \frac{2}{0.25 \, \text{m}} \sqrt{1 \, \text{kg} \times 9.8 \, \text{m/s}^{2}} = 25.0 \, \text{m/s}$$

For a grapefruit

$$m = \rho V = \rho \frac{4}{3} \pi r^{3} = (1000) \frac{4}{3} \pi (0.05)^{3} = 0.524 \, \text{kg}$$ $$v_{t} = \frac{2}{0.10 \, \text{m}} \sqrt{0.524 \, \text{kg} \times 9.8 \, \text{m/s}^{2}} = 45.8 \, \text{m/s}$$

Answer: $25.0 \, \text{m/s}$; $45.8 \, \text{m/s}$.

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