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SIZIF [17.4K]
2 years ago
5

A skydiver weighing 200 lbs with clothes that have a drag coefficient of .325 is falling in an area that has an atmospheric dens

ity of 1.225 kg/m2 (and assuming that altitude has a negligible effect on atmospheric density). The skydiver can change the body orientation from straight-erect with a cross-sectional area of 6 sqft to a belly-flat cross-sectional area of 24 sqft. Calculate the terminal velocity of the person when the body has straight and when the body has belly-flat orientations. Calculate the terminal velocity on these two different orientations.
Physics
1 answer:
TiliK225 [7]2 years ago
3 0

Answer:

The right solution is:

(a) 89.455 m/s

(b) 44.73 m/s

Explanation:

The given values are:

Mass,

m = 200 lbs

or,

   = \frac{200}{2.205} \ kg

   = 90.7 \ kg

Air's density,

\delta = 1.225 \ kg/m^3

Drag coefficient,

C_d=0.325

When body is straight, area,

A_1=6 \ ft^2

As we know,

Terminal velocity,

⇒  V_t=\sqrt{\frac{2W}{C_d \delta A} }

or,

⇒      =\sqrt{\frac{2mg}{C_d \delta A} }

At straight orientation,

⇒ V_t'=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558} }

⇒      =\sqrt{\frac{1777.72}{0.223}}

⇒      =89.455 \ m/s

When belly flat,

⇒  V_t''=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558\times 4} }

⇒        =\sqrt{\frac{1777.72}{0.889} }

⇒        =44.73 \ m/s

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