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
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