Answer
$h(t) = -4.9t^2 + 1800$
$19.166$ seconds
Work Step by Step
To find the height function, start with the acceleration function. Since the acceleration due to gravity close to Earth is $-9.8 m/s^2$, the acceleration function is:
$a(t) = -9.8$
Integrate this to find the velocity function.
$\int a(t) = v(t) = -9.8t + v_0$
Since the rock is dropped, the initial velocity of the rock is zero.
$v_0 = 0$
$v(t) = -9.8t$
Integrate this to find the height function.
$\int v(t) = h(t) = -4.9t^2 + h_0$
Since the rock is dropped from 1800 meters above the canyon floor, the initial height of the rock is 1800.
$h_0 = 1800$
$h(t) = -4.9t^2 + 1800$
The rock will hit the canyon floor when the height is zero. Therefore, to find the time at which the rock hits the canyon floor, set $h(t) = 0$.
$0 = -4.9t^2 + 1800$
$4.9t^2 = 1800$
$t^2 = 367.347$
$t = 19.166$ seconds
