Use a(t) = −9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.) A canyon is 1300 meters deep at its deepest point. A rock is dropped from the rim above this point. How long will it take the rock to hit the canyon floor? (Round your answer to one decimal place.)

t = 16.3 s

Step-by-step explanation:

The equation to determine the time it will take to get to the canyon floor is.

H = ut - 1/2(gt²)

In this case

U = initial velocity= 0

H = 1300 metres

g = -9.8 ms^-2

1300= 0 - 1/2(-9.8t²)

1300= 9.8t²/2

1300*2= 9.8t²

2600= 9.8t²

2600/9.8= t²

265.306= t²

√265.306 = t

16.288 =t

To one decimal place

t = 16.3 s