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. 2-34).

(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*t-1/2 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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