Posted by colin on September 09, 2002 at 11:16:48:
Here's the problem:
A body with mass m is projected vertically downward with initial velocity v0 in a medium offering resistance proportional to the square root of the magnitude of the velocity. Find the relation between the velocity v and the time t.
The answer given in the book to this is:
2m/k(sqrt[v0]-sqrt[v]) + 2m^2g/k^2[mg-kSqrt[v0]/mg-kSqrt[v] = t
Now how the heck did they get that? I've been practically killing myself trying to understand this, and still can't.
I set up the problem as:
dv/dt = g - k/m*Sqrt[v]
dt = dv/(g-c*Sqrt[v]) where c = k/m
t = Int[dv/(g-c*Sqrt[v])] + c
from this I can solve for t with 1 integration. but the solution they give looks like 2 integrations were done...? why? can someone explain?
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