Equation of Motion of sky-diver

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SUMMARY

The equation of motion for a free-falling sky-diver can be expressed as dv/dt = -g + (k/M) v^2, where v represents the sky-diver's velocity, k is the air resistance proportionality constant, and g is the gravitational constant. The solution to the differential equation yields the velocity function v(t) = √((Mg/k) ((e^(-At) - 1)/(e^(-At) + 1))), with A defined as 2√(gk/M). As time approaches infinity, the terminal velocity vL is established as vL = √(mg/k), indicating that the sky-diver reaches a constant speed due to the balance of gravitational and air resistance forces.

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  • Understanding of differential equations
  • Familiarity with concepts of gravitational force and air resistance
  • Knowledge of limits and asymptotic behavior in calculus
  • Basic understanding of the physics of free fall
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Samantha24
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Homework Statement



A free-falling sky-diver of mass M jumps from an aeroplane and beforen he opens his parachute experiences air resistance which is proportional to the square of the magnitude of his velocity.

a) Show that the equation of motion for the sky-diver can be written as
dv/dt = -g + (k/M) v^2

where v is the magnitude of the sky-diver's velocity, k is the proportionality constant for the air resistance and g is the gravitational constant.

b) What initial data concerning, v(0), is consistent with the following solution
the of differential equation obtained in part (a) of this question,

v(t) = square root ((Mg/k) ((e^-At - 1/ e^-At + 1)) ;

where
A = 2* (square root gk/M)

c) Using the expression for the velocity obtained in part (b) of this question, show that there exists a limiting or terminal velocity, vL, such that

v(t) tends to vL, t tends to infinity.

What is vL?

Homework Equations



dv/dt = -g + (k/M) v^2

v(t) = square root ((Mg/k) ((e^-At - 1/ e^-At + 1)) ;

where
A = 2* (square root gk/M)

The Attempt at a Solution



a) done

b)

dv/[(k/m)v^2 - g] = 1
sqrt(m/kg)arctan(v sqrt k/mg) = t + c
arctan(v sqrt k/mg) = sqrt(kg/m)*(t+C)
v sqrt(k/mg) = tan[sqrt(kg/m)*(t+C)]
v(t) = sqrt(mg/k)tan[sqrt(kg/m)*(t+C)]

c)

vL = lim{t→∞} √(mg/k)*(e^(-At) - 1)/(e^(-At) + 1) = √(mg/k)

I can't get the exponentials. Help appreciated.
 
Last edited:
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Hi Samantha! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

Hint: what is the limit as t -> ∞ of (e-At - 1)/(e-At + 1) ? :wink:
 


Hey Tim :smile:

Thanks! I got it!
 

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