Conservation of Mechanical Energy Question!

Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 8.5-kg mass m1 is released and falls through a vertical distance of h = 1.83 m before hitting the ground. Use conservation of mechanical energy to determine:

1. The velocity of m1 just before m2 hits the ground.
2. The maximum height attained by m2.

What I've tried:
PE + KE = E
mgy + .5mgv^2 = E
Before the system accelerates, KE = 0, so mgy = E.
(8.5*9.8*1.83) = 152.44 J

That's pretty much the only thing I'm certain of.
I've tried setting KE = 152.44 to solve for the velocity, but that is incorrect.
I've tried setting 152.44 = mgy of m2, but the new y is incorrect.
...and a whole bunch of variations of this sort of thing, using various help tools, all incorrect.

Can someone please show me what I'm supposed to be doing, using the actual numbers in this problem? I'm afraid to have a theoretical answer because every one I've tried has failed.

Thanks in advance.

(Hmmm...It says that "m1 hits the ground", but then part (1) says, "...just before m2 hits the ground." So which is it? A link to "the figure" sure would have been helpful.)

What the conservation of mechanical energy really says is that the sum of the PE and the KE is _constant_; that is it's the same "before" the weight drops and "after" the weight drops (and just before it hits the ground):

PE_before + KE_before = PE_after + KE_after

That's really the equation you should be using. Also, you need to include the PE and KE for both m1 and m2 in the equation.

So, start with the total energy "before":
PE_before = (m1)g(y1_before) + (m2)g(y2_before)
KE_before = 0

So, Total energy "before" is:
E_before = (m1)g(y1_before) + (m2)g(y2_before)