Posted by Amedus on Thursday, September 23, 2010 at 7:07pm.
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?

Physics  bobpursley, Thursday, September 23, 2010 at 7:17pm
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.

Physics  Anonymous, Tuesday, September 15, 2015 at 8:49pm
5.05s
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