# Launch with Chains

benf.stokes
I've read the FAQ and this is not homework, it is just a hard problem I can't solve.

A mass M attached to an end of a very long chain of mass per unit length $$\lambda$$
, is thrown vertically up with velocity $$v_{0}$$.
Show that the maximum height that M can reach is:

$$h=\frac{M}{\lambda}\cdot \left [ \sqrt[3]{1+\frac{3\cdot \lambda\cdot v_{o}^{2}}{2\cdot M\cdot g}}-1 \right ]$$

and that the velocity of M when it returns to the ground is $$v=\sqrt{2\cdot g\cdot h}$$
Conservation of energy cannot be used because inelastic collisions occur in bringing parts of the rope from zero velocity to v

I start by setting up that the total mass at a position y is:
$$M_{total}=M+\lambda\cdot y$$ and thus the momentum at any position is given by:

$$p=(M+\lambda\cdot y)\cdot v$$ but I can't figure out an expression for v and using

$$F=\frac{dp}{dt}$$ I get an differential equation I can't solve.

Any help would be appreciated.

Homework Helper
hi benf.stokes!
Conservation of energy cannot be used because inelastic collisions occur in bringing parts of the rope from zero velocity to v

No, you can use conservation of energy …

the question asks for the maximum height, which would be in the limiting case of no energy loss.

As for the question itself, I don't understand what position the mass and chain are in at the start.

benf.stokes
Hi tiny-tim. I think that conservation of energy can't be used is a part of the problem,my bad, and if you used conservation of energy wouldn't you get at most a square root and not a cubic one?
At the start both the mass and the chain are at rest on the floor. Thanks for the reply

benf.stokes
i already arrived at the solution. Thanks anyway tiny-tim