A skydiver of mass \(82.3 \mathrm{~kg}\) (including outfit and equipment) floats downward suspended from her parachute, having reached terminal speed. The drag coefficient is 0.533 , and the area of her parachute is \(20.11 \mathrm{~m}^{2} .\) The density of air is \(1.14 \mathrm{~kg} / \mathrm{m}^{3}\). What is the air's drag force on her?

The drag force formula is given by the equation: F_d = 0.5 * C_d * ρ * A * v^2 where F_d is the drag force, C_d is the drag coefficient, ρ( rho) is the air density, A is the object's reference area (parachute area in this case), and v is the velocity of the object. Since the skydiver has reached terminal speed, the drag force equals the gravitational force acting on her. Therefore, we can use Newton's second law of motion to write this relationship: F_d = m * g where m is the mass of the skydiver, and g is the gravitational acceleration (approximately 9.81 m/s^2). We will use these equations to find the drag force.

We can now set the terminal speed drag force equal to the gravitational force: 0.5 * C_d * ρ * A * v^2 = m * g We have all the values except the velocity (v), so we'll first find the terminal velocity. Rearranging the equation for v: v^2 = (2 * m * g) / (C_d * ρ * A) Now, we can plug in the given values: v^2 = (2 * 82.3 * 9.81) / (0.533 * 1.14 * 20.11) Calculating the velocity: v^2 ≈ 127.14 v ≈ sqrt(127.14) ≈ 11.28 m/s Now that we have the terminal velocity, we can calculate the drag force using the drag force formula: F_d = 0.5 * C_d * ρ * A * v^2 F_d = 0.5 * 0.533 * 1.14 * 20.11 * (11.28)^2 Calculating the drag force: F_d ≈ 807.69 N So, the air's drag force on the skydiver is approximately 807.69 Newtons.