An 85 kg skydiver falls spread eagle with an estimated frontal area of 0.70 square meters, a drag coefficient of 1.0, and air density of 1.21 kilograms per cubic meter. Using g = 9.80 m/s squared, what terminal velocity does this give?

1. At terminal velocity, drag equals weight: mg equals one half times rho times C times A times v squared. 2. Solving for v gives v equal to the square root of (2mg divided by (rho times C times A)). 3. The numerator is 2 times 85 kg times 9.80 m/s squared, which is 1666. 4. The denominator is 1.21 times 1.0 times 0.70, which is 0.847. 5. Dividing gives about 1967, and taking the square root gives about 44 m/s. 6. 88 m/s would come from forgetting to take the square root of the ratio, and 22 m/s or 11 m/s would come from dividing the correct answer by factors of two rather than recomputing. _Source: OpenStax College Physics (CC BY 4.0), Ch 5 "Further Applications of Newton's Laws: Friction, Drag, and Elasticity", section 5.2 Drag Forces_