(adsbygoogle = window.adsbygoogle || []).push({}); Tough question -- tough ODE

1.I'm working over various texts this summer, one being Kibble's 'Classical Mechanics'. This being problem 15 on page 45. I'm stuck and your help would bemuch appreciated!!

Q) A particle moves vertically under gravity and a retarding force proportional to the square of its velocity.

2.The relevant equation of motion is this:

[tex]\ddot{z} = -g -k\dot{z}^2[/tex] (k is a constant, g grav. acceleration)

What I need to work out is its position at time, t, i.e. [tex]z(t)[/tex].

3.If the force were proportional to the velocity, and not the square of it, I would integrate once and then the resulting ODE i'd solve by finding some integrating factor.

However, I have issues with the [tex]\left(\frac{dz}{dt}\right)^2[/tex] term.

One thing Ithoughtwould fix this would be the substitution,

[tex]p = \left(\frac{dz}{dt}\right)[/tex]

This yields the equation,

[tex]pdp + (kp^2 + g)dz = 0 [/tex]

which I made exact by multiplying through by

[tex]e^{2kz}[/tex]

which is then solved by integration -- yields,

[tex] \frac{p^2e^{2kz}}{2} + \frac{g}{2k}e^{2kz} = c [/tex]

where c is a constant.

Rearranging forpthen and since

[tex]p = \left(\frac{dz}{dt}\right)[/tex]

we can find z:

[tex] \left(\frac{dz}{dt}\right) = \sqrt{2ce^{-2kz} - g/k}[/tex]

The GIVEN ANSWER is:

[tex]z = z_0 + \frac{1}{k}\ln\cos[\sqrt{gk}{(t_0 - t)}][/tex]

It's quite possible that [tex]c = -z_0[/tex] if i remember rightly, but I still cannot see how I'm going to get to the answer from here; rather in fact, how i might adequately find z(t) at all.

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# Homework Help: Tough question - tough ODE

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