# another linear motion question

• Aug 14th 2010, 09:39 AM
FGT12
another linear motion question
A particle moves vertically under gravity and a retarding force proportional to the square of the velocity. If $v$ is its upward or downward speed and we know that $\dot{v}=-g-kv^2$, show that its position at time t is given by
$z = z_0 + \frac{1}{k}ln\cos[\sqrtgk (t_0 -t)]$
• Aug 14th 2010, 10:37 AM
Ackbeet
What have you done so far?
• Aug 14th 2010, 11:58 AM
FGT12
I integrated with respect with time twice
$\frac{dv}{dt}=-g-kv^2$
$v= v_0 - gt -kv^2t$
$x= x_0 + v_0t -\frac{gt^2}{2} -\frac{-kv^2t^2}{2}+$

however i feel there is a technique to solce DEs that i dont know, because i cannot see where the ln cos... term would come from
• Aug 14th 2010, 02:59 PM
Ackbeet
Your technique for solving the DE is, alas, entirely wrong. The reason is that the function v occurs in the RHS. You can't just integrate with respect to t, and treat v like a constant, when it varies with t. This equation is separable. Try

$\displaystyle{\int\frac{dv}{-g-kv^{2}}=\int dt}.$
• Aug 16th 2010, 03:12 AM
FGT12
I integrate the above and get
$\frac{-1}{k}\int \frac{1}{v^2 + \frac{g}{k}} dv = \int dt$
$\frac{-1}{\sqrt (gk)}arctan (\frac{\sqrt (k)}{\sqrt(g)}v) = t - t_0$
$arctan(\frac{\sqrt (k)}{\sqrt (g)}v)= \sqrt (gk) (t_0 -t)$

however from here taking the tangent of both sides does not leave anything readily integrable
• Aug 16th 2010, 03:50 AM
Ackbeet
Quote:

however from here taking the tangent of both sides does not leave anything readily integrable
I think it does. You basically have the tangent of a linear function of t. So the numerator looks like the derivative of the denominator. What does that give you?
• Aug 16th 2010, 03:55 AM
FGT12
thanks
• Aug 16th 2010, 03:56 AM
Ackbeet
You're welcome. Have a good one!