A coin dropped from the top of a precipice (cliff) takes 7 s to hit the ground. How high is the precipice (cliff)?

Question

kkutie7 · Answer

use this equation
$$y=y_{0}+V_{0}+\frac{1}{2}at^{2}$$

kkutie7 · Answer

y is the height
y0 is initial position 
V0 is the initial velocity
a is the acceleration
t is the time

kkutie7 · Answer

you want to find why.
y0=0 considering that you are starting at the origin (0,0)
V0=0 we can assume the coin is being dropped from rest
a= 9.8 m/s^2 gravity 
and t is the time given
plug in the values and solve for y