physics
posted by hana on .
A stone is thrown vertically upward with a speed of 11.5 m/s from the edge of a cliff 95.0 m high.
(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?
For (c), the right answer is 108.46 m. I really need help on parts (a) and (b) please!

(a) Solve this for time t:
Y = 11.5 t 4.9 t^2 = 95
(b) V^2(final)  V^2(initial) = 2 g H
(Easily derived using conservation of energy)
H is the distance the stone falls, 95 m
(c) You said you got this already. It would be twice the distance it goes up (since it comes back down), plus 95m
Total distance = (V^2/g) + H