Answer:
1.75 s
Explanation:
The vertical position of the rock at time t is given by:
[tex]y(t) = h +ut - \frac{1}{2}gt^2[/tex]
where
h = 15.0 m is the initial height
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration of gravity
t is the time
When the rock reaches the bottom of the cliff,
y(t) = 0
Substituting this condition into the previous equation, we can find the time at which the rock reaches the bottom of the cliff:
[tex]0 = h - \frac{1}{2}gt^2\\t= \sqrt{\frac{2h}{g}}=\sqrt{\frac{2(15.0)}{9.8}}=1.75 s[/tex]
