A rock is thrown vertically upward from the edge of a cliff. The rock reaches a maximum height of 18.6 m above the top of the cliff before falling to the base of the cliff, landing 6.60 s after it was thrown. How high is the cliff? Assume gravity= 9.8m/s2
Vyf=Vyo + at
Y=Yo + 1/2(Vyo+Vyf)t
Y=Yo + Vyot + 1/2at2
The Attempt at a Solution
I don't know where to start. I must figure out initial speed and the difference between the cliff and the max height. Is the initial position 0 at the height of the cliff or at the max height? Do I care that when the ball reaches the max height the speed is 0?
EDIT:Figured it out, I had the dumb for few minutes