A free-falling sky-diver. Using ODE and Variable Separation

[Richard Platt]

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

Within the Question

The Attempt at a Solution

I'm stuck on (b).

dv/(kv^2-Mg)= dt/M

Intregrate both sides;

(1/2kv) ln |kv^2-Mg|= t/m + c?

Then what to do..? Something to do with with Ae^(-kt/M)..?

 

[anathema]

To solve for the initial velocity, we can set t=0 and solve for v(0).

(1/2kv(0)) ln |kv(0)^2-Mg|= 0 + c

c= (1/2kv(0)) ln |kv(0)^2-Mg|

Now, we can plug in the given solution for v(t) into the equation and solve for v(0).

v(t) = sqrt((Mg/k)(e^-At - 1/ e^-At + 1))

v(0) = sqrt((Mg/k)(e^0 - 1/ e^0 + 1))

v(0) = sqrt((Mg/k)(1 - 1 + 1))

v(0) = sqrt(Mg/k)

Therefore, the initial velocity, v(0), is sqrt(Mg/k).