Question

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

Answer 1

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,

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

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

⇒        $=44.73 \ m/s$