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?
hf=hi+Vi*t-1/2 g t^2
you know hf=-59
hi=0 solve for t, notice it is a quadratic.
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.