Physics
posted by Hannah .
At t = 0, a stone is dropped from a cliff above a lake; 2.0 seconds later another stone is thrown downward from the same point with an initial speed of 33 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

I have just recently had to answer a very similar question. My prof. has given me the two general eqns to use:
Y(final)=Y(initial)+V(initial)((T(final)T(initial))+1/2a(T(final)T(initial))^2
and
V(final)=V(initial)+a(T(final)T(initial))
From there, write down your knowns and derive more specific eqns from the general ones... Very vague i know...good luck 
Y1 = (g/2) t^2
Y2 = 33(t2) (g/2)((t2)^2
Set Y1  Y2 and solve for t.
Use that t to compute Y at that time. 
when i solve for t I get an extremely small number like .1998. I don't think I am setting up the problem right or something.