(v_o v_t)/(v_0^2 + v_t^2)^(1/2)

Solution:
  The equation of motion for the upward motion is

m ^2x/t^2 = -m k v^2 - m g (1)

where k is constant of proportionality. Using the relation ^2x/t^2 = v/t = v/x x/t=v v/x
we can write (1) as

(v   v)/(k v^2 + g) = -x (2)

Integrating (2), we find

log(k v^2 + g)/(2 k) = c_1 - x (3)

and using the initial condition of v = v_0 when x = 0, we find

c_1 = log(k v_0^2 + g)/(2 k) (4)
∴ x = 1/(2 k) log((k v_0^2 + g)/(k v^2 + g)) (5)

Now, the equation of the downward motion is (using the same coordinate system):


m ^2x/t^2=m k v^2-m g		 (6)
(v v)/(g - k v^2) = -x (7)

Integrating (7)  , we get

-x = c_2 - log(g - k v^2)/(2 k) (8)

and requiring v=0 at x_ (max height) gives

-x_ (max height) = c_2 - log(g)/(2 k) (9)

Solving (5) for the max height (when v=0), we find

x_ (max height) = log((k v_0^2 + g)/g)/(2 k) (10)

and plugging (10) into (9) gives

c_2 = log(g^2/(k v_0^2 + g))/(2 k) (11)

and plugging (11) into (8) gives

x = 1/(2 k) log(g - k v^2) - 1/(2 k) log(g^2/(k v_0^2 + g)) (12)
or, x = 1/(2 k) log(((g - k v^2) (k v_0^2 + g))/g^2) (13)

The mass will have returned to it's initial position when x=0, so

g^2 = (g - k v^2) (k v_0^2 + g) (14)
k v^2 = g - g^2/(k v_0^2 + g) (15)
v^2 = g/k - g^2/(k^2 v_0^2 + g k) (16)

We can solve for the terminal velocity by equating the retarding force to the gravitational force:
m g=m k v_t^2
or
v_t=g/k^(1/2)  (Again note, the form of the terminal velocity is decidedly unphyiscal, because it appears to be independent of the mass of the object. This            is an unphysical consequence of the postulated force -- the terminal velocity should depend on the mass. )

so (16) becomes

v^2 = v_t^2 - v_t^4/(v_0^2 + v_t^2) (17)

Or

v = (v_t v_0)/(v_0^2 + v_t^2)^(1/2) (18)
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