## Drag Coefficient

Consider during a round of golf, a player smacks the 0.5 lb ball such that it travels 200 ft/s. If the drag force acting on the 0.1 ft^2 projected area is 2.5 lbf, what is the drag coefficient? Assume the density of air is 0.08 lb/ft^3.

Hint
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$D=\frac{1}{2}\rho U^{2}C_{D}A$$$where $$C_D$$ is the drag coefficient, $$U$$ is the flowing fluid or moving object’s velocity, $$\rho$$ is the fluid density, and $$A$$ is the projected area of blunt objects with axes perpendicular to the flow. Hint 2 $$1\:lbf=1\:lb \times gravity\approx 1\:lb \times 32.2 \frac{ft}{s^2}$$$
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$D_f=\frac{1}{2}\rho U^{2}C_{D}A$$$where $$C_D$$ is the drag coefficient, $$U$$ is the flowing fluid or moving object’s velocity, $$\rho$$ is the fluid density, and $$A$$ is the projected area of blunt objects with axes perpendicular to the flow. $$2.5lb_f=\frac{1}{2}(0.08\frac{lb}{ft^3})(200\frac{ft}{s})^2C_D(0.1ft^2)$$$
Before solving for $$C_D$$ , recall that $$1\:lb_f=1\:lb \times gravity\approx 1\:lb \times 32.2 \frac{ft}{s^2}$$ . Thus,
$$2.5lb\times32.2\frac{ft}{s^2}=\frac{1}{2}(0.08\frac{lb}{ft^3})(40,000\frac{ft^2}{s^2})C_D(0.1ft^2)$$$$$80.5=(160)C_D$$$
$$C_D=\frac{80.5}{160}=0.503\approx 0.5$$\$
0.5