Question

A falling body on Earth will generally not fall at constant acceleration in reality, due to air resistance (or drag). The equation for the drag force Fd is Fd = 0.5 (Cd A) ρ v2 where Cd is the unitless drag coefficient of a body, A is its cross-sectional area, ρ is the density of the air, and v is the body's velocity with respect to the atmosphere.(b) Look up and then see if you can also calculate the terminal velocity (that is, the final nearly constant speed) that a falling, parachuteless person would have after diving from, let's say, a few kilometers up.

Answer 1

at terminal velocity,

weight = drag force

the question has not given the mass of person, so I will assume all the values.

let mass of person, m = 75 kg

Please note here the procedure matters , not the values of mass, Concentrate on procedure

Without parachute, a person will most probably fall in head down position.

so, In this position, A = 0.18 m² , C_d = 0.7, = 1.21 kg/m³ ( I have looked up these values from online and textbook resources)

so,

mg = 1/2v²AC_d

v² = 2mg / AC_d

v = sqrt ( 2mg / AC_d)

v = sqrt ( 2 * 75 * 9.8 / 1.21 * 0.18 * 0.7)

v = 98.2 m/s