Complete Question
For a human body falling through air in a spread edge position , the numerical value of the constant D is about [tex]D = 0.2500 kg/m[/tex]
What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures?
Answer:
The value of D is [tex]D = 0.457 \ kg/m[/tex]
Explanation:
From the question we are told that
The terminal velocity is [tex]v_t = 42.7 \ m/s[/tex]
The mass of the skydiver is [tex]m = 85.0 \ kg[/tex]
The numerical value of D is [tex]D = 0.2500 kg/m[/tex]
From the unit of D in the question we can evaluate D as
[tex]D = \frac{m * g }{v^2}[/tex]
substituting values
[tex]D = \frac{85 * 9.8 }{(42.7)^2}[/tex]
[tex]D = 0.457 \ kg/m[/tex]
