Answer:
The time it takes the rock to reach the canyon floor is approximately 4 seconds.
Step-by-step explanation:
The equation representing the height h (in feet) of an object t seconds after it is dropped is:
[tex]h=-16t^{2}+h_{0}[/tex]
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:
[tex]h=-16t^{2}+h_{0}[/tex]
[tex]0=-16t^{2}+255\\\\16t^{2}=255\\\\t^{2}=\frac{255}{16}\\\\t^{2}=15.9375\\\\t=\sqrt{15.9375}\\\\t=3.99218\\\\t\approx 4[/tex]
Thus, the time it takes the rock to reach the canyon floor is approximately 4 seconds.
