Physics
posted by Amedus on .
A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 85.0 m high (Fig. 234).
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?

It is the same equations as always:
d=Vi*t1/2 g t^2
Vf=Vi+gt
Now for total distance, you will have to calculate the height it reaches at the top (when vf is zero). Double that, add to height of cliff. 
5.05s