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*t1/2 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=Vigt solve for t at the top.
h=Vi*t1/2 g t^2 solve for h, heightabove clift.
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