Physics
posted by Carden on .
A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 27 m high. When it hits the ground at the base of the cliff the rock has a speed of 24 m/s.
(a) Assuming that air resistance can be ignored, find the initial speed of the rock.
(b) Find the greatest height of the rock as measured from the base of the cliff.
I know I have to use projectile motion somehow. I don't know how to without knowing the initial velocity. I can't find that without solving the first part. I'm having trouble picking an equation. Please help.

Initial KE+ initialPE=final KE
I suspect you can solve it from that.
Greatest height? Final KE= greatest PE
solve for greatest height. 
KE initial = 0
PE initial = mgh = 68.8662 = KE final
KE final = 1/2mv2
v = 23.01
That isn't right. What have I done wrong? 
Where did you get initial KE is zero? That is what you are solving for. You need a tutor, pronto.

I am just so confused by this problem. I have done other problems with ease and I'm having so much trouble with other ones.

you do:
1/2mv(initial)^2+mgh=1/2mv(final)^2
so simplify that to : v(inital)= sqrt(v(final)^22gh)
For b): 1/2mv(initial)^2=mgh
solve for h : h=v^2/2g
then add your original height to the new height for your answer