help me with this math..?

Question

the question. solve m(dv/dt) = mg - kv^2 .. (ordinary differential eq.)

then.. a man and a parachute are falling with a speed of 173 ft/sec. when the parachute opens and the speed is reduced so as to approach the limiting value of 15 ft/sec. by air resistance proportional to the square of the speed.

show that

t = (15/2g)ln((79/94).[(v+15)/(v-15)])

i have tired thinking.. still cant show that.. :(

Answer 1

I see no one try to help you after 6 hours...
m(dv/dt) = mg - kv^2, or
dt = m dv/(mg - kv^2) = (1/g)dv /{1 - (k/mg)v^2}
= (1/2g)dv{1/[1 - (k/mg)^0.5v] + 1/[1 + (k/mg)^0.5v]}
= (1/2g)[(mg/k)^0.5]d[(k/mg)^0.5v]{1/[1 - (k/mg)^0.5v] + 1/[1 + (k/mg)^0.5v]}
= 0.5(m/kg)^0.5d[(k/mg)^0.5v]{1/[1 - (k/mg)^0.5v] + 1/[1 + (k/mg)^0.5v]}
After integration we get:
t + C =
= 0.5(m/kg)^0.5 Ln{[1 + (k/mg)^0.5v]/ [1 - (k/mg)^0.5v]}
Or: [1 + (k/mg)^0.5v]/ [1 - (k/mg)^0.5v] = C'*exp[2t(kg/m)^0.5]

For the wording problem, I think that the differential eq. is the right one. However, the wording problem is incomplete. For example, we do not know for the air resistance the proportional constant to the square of the air speed. I believe that you know all the parameters in the equation I have diduced. Good Luck!