# Question #3d8fc

Aug 24, 2017

$177$ $\text{m}$

#### Explanation:

As the ball is "thrown vertically down from the edge" of the cliff, the distance fallen will be the height of the cliff.

Let's use the formula $s = u t + \frac{1}{2} a {t}^{2}$; where $s$ is the distance, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time taken.

In this case, $a$ will be equal to $9.81$ ${\text{m/s}}^{2}$, which is the acceleration due to gravity:

$R i g h t a r r o w s = 0$ $\text{m/s}$ $\times 6$ $\text{s}$ $+ \frac{1}{2} \times 9.81$ ${\text{m/s}}^{2} \times {6}^{2}$ ${\text{s}}^{2}$

$R i g h t a r r o w s = 4.905$ ${\text{m/s}}^{2} \times 36$ ${\text{s}}^{2}$

$\therefore s = 176.58$ $\text{m}$

Therefore, the cliff is around $177$ $\text{m}$ high.