To solve this problem it is necessary to apply the concepts related to the Drag Force and the Weight of a body described since Newton's second law.
The drag force is given as
[tex]\frac{1}{2} \rho C_D v^2 A = W[/tex]
Where
[tex]\rho[/tex]= Density
[tex]C_d[/tex]= Drag Coefficient
v = Velocity
A = Cross-sectional Area
W = Weight
Replacing to find the velocity we have that
[tex]\frac{1}{2} \rho C_D v^2 A = W[/tex]
[tex]\frac{1}{2} (2.38*10^{-3})(1)v^2 (200) = 200[/tex]
[tex]V = 28.98ft/s[/tex]
Converting these values to miles per hour we have to
[tex]V = 28.98ft/s(\frac{3600s}{1hour})(\frac{1mile}{5280ft})[/tex]
[tex]V = 19.76mi/h[/tex]
Therefore the highest downward speed that this jumper can attain is 19.76mi/h
