Question

a stone is thrown vertically upward with a speed of 16.0 m/s from the edge of a cliff 75.0 m high. how much later does it reach the bottom of the cliff?

Answer 1

Below

Step-by-step explanation:

df = do + vo t + 1/2 a t^2 df = final position = ground = 0

do - original position = 75 m

vo = 16 m/s

a = -9.81 m/s^2

0 = 75 + 16 t - 1/2 ( 9.81 )t^2

- 4.905 t^2 + 16 t + 75 = 0 Use Quadratic Formula to find :

t = 5.9 s