Question

Use a(t) = −9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.) A canyon is 2100 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:

$t \approx 20.7\,s$

Step-by-step explanation:

Given that rock experiments a constant acceleration, position function can be obtained by integrating twice:

$v(t) = v_{o} - a\cdot t$

$s(t) = s_{o} + v_{o}\cdot t -\frac{a}{2}\cdot t^{2}$

The initial conditions of the rock are, respectively:

$s_{o} = 2100\,m$

$v_{o} = 0\,\frac{m}{s}$

Position of the rock as a function of time is:

$s(t) = 2100\,m -(4.9\,\frac{m}{s^{2}})\cdot t^{2}$

The time taken for the rock to hit the canyon floor is:

$0\,m = 2100\,m - (4.9\,\frac{m}{s^{2}} )\cdot t^{2}$

$2100\,m = (4.9\,\frac{m}{s^{2}} )\cdot t^{2}$

$t = \sqrt{\frac{2100\,m}{4.9\,\frac{m}{s^{2}} } }$

$t \approx 20.7\,s$