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Provide an Appropriate Response mm Falling from Rest Under the Action of Gravity Encounters an Falling

Question 65

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Provide an appropriate response.
-If a body of mass mm falling from rest under the action of gravity encounters an air resistance proportional to three times the square root of velocity, then the body's velocity tt seconds into the fall satisfies the equation: mdvdt=mg3kv2\mathrm { m } \frac { \mathrm { dv } } { \mathrm { dt } } = \mathrm { mg } - 3 \mathrm { kv } ^ { 2 } where k\mathrm { k } is a constant that depends on the body's aerodynamic properties and the density of the air.
Determine the equilibrium, velocity curve, and the terminal velocity for a 150lb150 \mathrm { lb } skydiver (mg=150) ( \mathrm { mg } = 150 ) with k=0\mathrm { k } = 0 .


A) Equilibrium: v=mg3k\mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }
 Provide an appropriate response. -If a body of mass  m  falling from rest under the action of gravity encounters an air resistance proportional to three times the square root of velocity, then the body's velocity  t  seconds into the fall satisfies the equation:  \mathrm { m } \frac { \mathrm { dv } } { \mathrm { dt } } = \mathrm { mg } - 3 \mathrm { kv } ^ { 2 }  where  \mathrm { k }  is a constant that depends on the body's aerodynamic properties and the density of the air. Determine the equilibrium, velocity curve, and the terminal velocity for a  150 \mathrm { lb }  skydiver  ( \mathrm { mg } = 150 )   with  \mathrm { k } = 0 . A)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   B)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   C)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   D)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     v _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }
vterminal =100ft/sec\mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }

B) Equilibrium: v=mg3k\mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }
 Provide an appropriate response. -If a body of mass  m  falling from rest under the action of gravity encounters an air resistance proportional to three times the square root of velocity, then the body's velocity  t  seconds into the fall satisfies the equation:  \mathrm { m } \frac { \mathrm { dv } } { \mathrm { dt } } = \mathrm { mg } - 3 \mathrm { kv } ^ { 2 }  where  \mathrm { k }  is a constant that depends on the body's aerodynamic properties and the density of the air. Determine the equilibrium, velocity curve, and the terminal velocity for a  150 \mathrm { lb }  skydiver  ( \mathrm { mg } = 150 )   with  \mathrm { k } = 0 . A)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   B)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   C)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   D)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     v _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }
vterminal =100ft/sec\mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }

C) Equilibrium: v=3kmgv = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }
 Provide an appropriate response. -If a body of mass  m  falling from rest under the action of gravity encounters an air resistance proportional to three times the square root of velocity, then the body's velocity  t  seconds into the fall satisfies the equation:  \mathrm { m } \frac { \mathrm { dv } } { \mathrm { dt } } = \mathrm { mg } - 3 \mathrm { kv } ^ { 2 }  where  \mathrm { k }  is a constant that depends on the body's aerodynamic properties and the density of the air. Determine the equilibrium, velocity curve, and the terminal velocity for a  150 \mathrm { lb }  skydiver  ( \mathrm { mg } = 150 )   with  \mathrm { k } = 0 . A)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   B)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   C)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   D)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     v _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }
vterminal =100ft/sec\mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }

D) Equilibrium: v=3kmgv = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }
 Provide an appropriate response. -If a body of mass  m  falling from rest under the action of gravity encounters an air resistance proportional to three times the square root of velocity, then the body's velocity  t  seconds into the fall satisfies the equation:  \mathrm { m } \frac { \mathrm { dv } } { \mathrm { dt } } = \mathrm { mg } - 3 \mathrm { kv } ^ { 2 }  where  \mathrm { k }  is a constant that depends on the body's aerodynamic properties and the density of the air. Determine the equilibrium, velocity curve, and the terminal velocity for a  150 \mathrm { lb }  skydiver  ( \mathrm { mg } = 150 )   with  \mathrm { k } = 0 . A)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   B)  Equilibrium:  \mathrm { v } = \sqrt { \frac { \mathrm { mg } } { 3 \mathrm { k } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   C)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     \mathrm { v } _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }   D)  Equilibrium:  v = \sqrt { \frac { 3 \mathrm { k } } { \mathrm { mg } } }     v _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }
vterminal =100ft/secv _ { \text {terminal } } = 100 \mathrm { ft } / \mathrm { sec }

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