Respuesta :
Answer:
The ratio of the cross-sectional area inthe slower position to that in the faster position is [tex]\bf{6.37}[/tex]
Explanation:
The terminal velocity of any object is given by
[tex]v_{t} = \sqrt{\dfrac{2 mg}{C \rho A}}[/tex]
where '[tex]m[/tex]' is the mass of the object, '[tex]g[/tex]' is the acceleration due to gravity, '[tex]C[/tex]' is the drag coefficient, '[tex]\rho[/tex]' is the density and '[tex]A[/tex]' is the cross-sectional area of the object.
For spread-eagle position if [tex]A_{S}[/tex] be the cross-sectional area and for nosedive position if [tex]A_{N}[/tex] be the cross-sectional area, then from the above expression
[tex]&& v_{t}^{S} = \sqrt{\dfrac{2 mg}{C \rho A_{S}}}\\&and,& V_{t}^{N} = \sqrt{\dfrac{2 mg}{C \rho A_{N}}}[/tex]
where [tex]v_{t}^{S}~and~v_{t}^{N}[/tex]are the terminal velocities for spread-eagle position and nosedive position respectively.
Taking the ratio of both the velocities,
[tex]&& \dfrac{v_{t}^{N}}{v_{t}^{S}} = \sqrt{\dfrac{A_{S}}{A_{N}}}\\&or,& \dfrac{A_{S}}{A_{N}} = \dfrac{(v_{t}^{N})^{2}}{(v_{t}^{S})^{2}} = \dfrac{308^{2}}{122^{2}} = 6.37[/tex]
