Question #3d8fc

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Question #3d8fc

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Answer 1 · 2017-08-24T19:41:40

$177$ $177$ $m$ $"m"$

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 $s = u t + \frac{1}{2} a t^{2}$ $s = u t + frac(1)(2) a t^(2)$ ; where $s$ $s$ is the distance, $u$ $u$ is the initial velocity, $a$ $a$ is the acceleration, and $t$ $t$ is the time taken.

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

$\Rightarrow s = 0$ $Rightarrow s = 0$ $m/s$ $"m/s"$ $\times 6$ $times 6$ $s$ $"s"$ $+ \frac{1}{2} \times 9.81$ $+ frac(1)(2) times 9.81$ ${m/s}^{2} \times 6^{2}$ $"m/s"^(2) times 6^(2)$ $s^{2}$ $"s"^(2)$

$\Rightarrow s = 4.905$ $Rightarrow s = 4.905$ ${m/s}^{2} \times 36$ $"m/s"^(2) times 36$ $s^{2}$ $"s"^(2)$

$therefore s = 176.58$ $"m"$

Therefore, the cliff is around $177$ $"m"$ high.

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Question #3d8fc

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