Question

A skydiver will reach a terminal velocity when the air drag equals their weight. For a skydiver with high speed and a large body, turbulence is a factor. The drag force then is approximately proportional to the square of the velocity. Taking the drag force to be F_D=1/2}rho A v²) and setting this equal to the person’s weight, find the terminal speed for a person falling "spread eagle."

a) v_t = √{2W/rho A}
b) v_t = √{W/rho A}
c) v_t = √{W/rho A}
d) v_t = √{2W/rho}

Answer 1

To find the terminal speed for a person falling in a spread-eagle position, we need to set the drag force equal to the person's weight. Using the equation for drag force and equating it to the weight, we can solve for the terminal speed.

Step-by-step explanation:

In order to find the terminal speed (Ut) for a person falling in a spread-eagle position, we need to set the drag force equal to the person's weight. The drag force, (FD), is given by:

FD = 1/2 * rho * A * v^2

where rho is the air density, A is the cross-sectional area of the person facing the fluid, and v is the velocity. Setting the drag force equal to the person's weight, we have:

FD = mg

where m is the mass of the person and g is the acceleration due to gravity.

By equating the two equations, we can find the terminal speed, Ut:

mg = 1/2 * rho * A * Ut^2

Solving for Ut, we get:

Ut = sqrt(2mg / (rho * A))