Answer:
The time it takes the rock to reach the canyon floor is approximately 4 seconds.
Explanation:
The equation representing the height h (in feet) of an object t seconds after it is dropped is:
![h=-16t^(2)+h_(0)](https://img.qammunity.org/2021/formulas/mathematics/high-school/6wgy3fjafymmzd2wmsjgl3qnhdcqyt2wrz.png)
Here, h₀ is the initial height of the object.
It is provided that a small rock dislodges from a ledge that is 255 ft above a canyon floor.
That is, h₀ = 255 ft.
So, when the rock to reaches the canyon floor the final height will be, h = 0.
Compute the time it takes the rock to reach the canyon floor as follows:
![h=-16t^(2)+h_(0)](https://img.qammunity.org/2021/formulas/mathematics/high-school/6wgy3fjafymmzd2wmsjgl3qnhdcqyt2wrz.png)
![0=-16t^(2)+255\\\\16t^(2)=255\\\\t^(2)=(255)/(16)\\\\t^(2)=15.9375\\\\t=√(15.9375)\\\\t=3.99218\\\\t\approx 4](https://img.qammunity.org/2021/formulas/mathematics/high-school/le9jpmzso54fa8f8dl71vw7x5o96i6v22x.png)
Thus, the time it takes the rock to reach the canyon floor is approximately 4 seconds.