A rock is thrown vertically upward from the edge of a cliff. The rock reaches a maximum height of 20.6 m above the top of the cliff before falling to the base of the cliff, landing 6.54 s after it was thrown. How high is the cliff?

Answer 1

V^2 = Vo^2 + 2g*d.
Vo^2 = V^2--2g*d
Vo^2 = 0 - (-19.6)*20.6 = 403.76
Vo = 20.1 m/s. = Initial velocity.

V = Vo + g*t.
Tr = (V-Vo)/g = (0-20.1)/-9.8 = 2.05 s.
= Rise time.

Tr + Tf = 6.54 s.
2.05 + Tf = 6.54
Tf = 6.54 - 2.05 = 4.49 s. = Fall time
from 20.6 m above the cliff to Gnd.

Vo*t + 0.5g*t^2 = 20.6 + h.
0 + 4.9t^2 = 20.6 + h
h = 4.9*(4.49)^2 - 20.6 = 78.2 m.