Mechanics - Elastic springs and strings

[mr bob]

A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length \frac{5l}{3}. Show that if P is projected vertically downwards from A with speed \sqrt(\frac{3gl}{2}), P will come to instantaneous rest after moving a distance \frac{10l}{3}.

I thought about working all this out by finding the energies before and after the projection.

Before:-
KE =\frac{3gl}{4}
GPE = 0
EPE = 0

After:-
KE = 0
GPE = -(y - 5/3L)g where y is the full length of stretched string.
EPE = \frac{gl(Y- 5/3L)^2}{2L(Y- 5/3L)}

i can't seem to get the anwer using these, are they correct?

 

[FrogPad]

This seems like more of a physics related question. I'm sure if you post it in the physics section someone would help you out a little faster. I haven't taken differential equation in awhile, but this seems like a pretty straightforward differential equation question. I could definitely be wrong though