The terminal speed of a sky diver is 137 km/h in the spread-eagle position and 318 km/h in the nosedive position. assuming that the diver's drag coefficient c does not change from one position to the other, find the ratio of the effective cross-sectional area a in the slower position to that in the faster position.

\frac{A₁}{A₂} = 5.4

Explanation:

from the question we are given:

speed at spread eagle position (v₁) = 137 km/h

speed at nose dive position (v₂) = 318 km/h

First of all what is terminal velocity,

Terminal velocity is the maximum velocity that is attainable by an object falling through a fluid, and it occurs when drag force is equal to the downward force of gravity (weight)

air drag = downward force of gravity

C x A x V^{2} = mg   (C = drag coefficient)

therefore for the two positions it would be

C x A₁ x V₁^{2} = mg

C x A₂ x V^₂{2} = mg

therefore we now have

C x A₁ x V₁^{2} = C x A₂ x V^₂{2}

A₁ x V₁^{2} = A₂ x V₂^{2}

now substituting the values of the velocities we have

A₁ x 137^{2} = A₂ x 318^{2}

18769 A₁ = 101124 A₂

\frac{A₁}{A₂} = 101124 ÷ 18769

\frac{A₁}{A₂} = 5.4

this could be newton#'s law of cooling. the cooling curve is exponential