# Question #aeb96

Jan 29, 2018

$\therefore$ The height of the cliff is $62.145 m$

#### Explanation:

Let initial velocity be given by $' u '$ = $6.0 m {s}^{-} 1$,
and
Final velocity be given by $' v ' = 0 m {s}^{-} 1$.
$' t ' = 3.0 s$ is the time taken by the stone to reach the ground, and
$' a ' = 9.81 m {s}^{-} 2$ is the acceleration due to gravity.
Let $' s '$ be the distance covered.
According to the position- time relationship or the second equation of motion:

$s = u t + \frac{1}{2} a {t}^{2}$

$s = 6.0 \frac{m}{\cancel{s}} \times 3.0 \cancel{s} + \frac{1}{2} \times 9.81 \frac{m}{\cancel{s}} ^ 2 \times {\left(3.0\right)}^{2} {\cancel{s}}^{2}$

$s = 18 m + \frac{9.81 \times 9}{2} m$

$s = 18 m + 44.145 m$

$s = 62.145 m$ is the distance covered by the stone from top of the cliff to the bottom.

$\therefore$ The height of the cliff is $62.145 m$