Posted by **Anonymous** on Sunday, September 13, 2009 at 9:31pm.

A stone is thrown vertically upward with a speed of 10.0 m/s from the edge of a cliff 59.0 m high.

How much later does it reach the bottom of the cliff?

What is its speed just before hitting?

What total distance did it travel?

- physics -
**bobpursley**, Sunday, September 13, 2009 at 9:36pm
vertical position:

hf=hi+Vi*t-1/2 g t^2

you know hf=-59

hi=0 solve for t, notice it is a quadratic.

Finalspeed=Vi -g*t

Vi=10m/s

total distance=59+2*heightaboveclift

Not to figure the height above. At the top, Vf=0 (think about that).

Vf=Vi-gt solve for t at the top.

h=Vi*t-1/2 g t^2 solve for h, heightabove clift.

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