physics
posted by Anonymous on .
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?

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.