Question

Question #3d8fc

Answer 1

`177` `"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` `"m/s"^(2)`, which is the acceleration due to gravity:

`Rightarrow s = 0` `"m/s"` `times 6` `"s"` `+ frac(1)(2) times 9.81` `"m/s"^(2) times 6^(2)` `"s"^(2)`

`Rightarrow s = 4.905` `"m/s"^(2) times 36` `"s"^(2)`

`therefore s = 176.58` `"m"`

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