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## Homework Statement

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/s

^{2}

## Homework Equations

Vy

_{f}=Vy

_{o}+ at

Y=Y

_{o}+ 1/2(Vy

_{o}+Vy

_{f})t

Y=Y

_{o}+ Vy

_{o}t + 1/2at

^{2}

Vy

_{f}

^{2}=Vy

_{o}

^{2}+ 2a(y-y

_{o})

## 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

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