Question

A rock is dropped from a vertical cliff. The rock takes 6.00 s to reach the ground below the cliff. What is the height of the cliff? (take g=10.0 m/s2)

Answer 1

Answer:

180 m

Explanation:

The rock follows a free-fall motion - so the vertical distance covered can be found by using the equation

$h=\frac{1}{2}gt^2$

where

g = 10 m/s^2 is the acceleration due to gravity

t = 6.00 s is the time of the fall

Substituting these data, we find the height of the cliff:

$h=\frac{1}{2}(10 m/s^2)(6.00 s)^2=180 m$