Question

A rock is dropped from the edge of a cliff 15.0 m high. How long does it take to reach the bottom of the cliff?

Answer 1

1.75 s

Step-by-step explanation:

The vertical position of the rock at time t is given by:

`y(t) = h +ut - (1)/(2)gt^2`

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:

`0 = h - (1)/(2)gt^2\\t= \sqrt{(2h)/(g)}=\sqrt{(2(15.0))/(9.8)}=1.75 s`