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

Answer:

t = 12.8 s

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

$\Delta y = v_y t + \frac{1}{2}at^2$

$800 = 0 + \frac{1}{2}(9.8) t^2$

$800 = 4.9 t^2$

now we have

$t^2 = \frac{800}{4.9}$

$t = \sqrt{163.6}$

$t = 12.8 s$