Question

Use a(t) = −9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.) A canyon is 800 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.)

Answer 1

t = 12.8 s

Step-by-step explanation:

As we know that the deepest point of the canyon is 800 m

so here we will have the displacement of the rock in the canyon is same as that of depth of the canyon

So here we will say

`\Delta y = v_y t + (1)/(2)at^2`

`800 = 0 + (1)/(2)(9.8) t^2`

`800 = 4.9 t^2`

now we have

`t^2 = (800)/(4.9)`

`t = √(163.6)`

`t = 12.8 s`