The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. What is the height of the rock as a function of time, t in seconds? How long will it take the rock to hit the canyon floor?

In different treatments, I've seen #g# approximated by #-10#, #-9.8# or #-9.81# #m#/#s^2#.

The rock hits the canyon floor when #h=0#, so solve the resulting equation.

Acceleration due to gravity is #g#. I'll use #g ~~ 9.8# #m#/#s^2# in the negative direction.

Velocity at time #t# is the antiderivative of acceleration, so
#v(t) = -9.8 t+C#

#C# is the velocity when #t = 0# (the initial velocity). Since the stone was dropped, #C = 0#.

So #v(t) = -9.8 t#

Position (height) is the antiderivative of velocity.

#h(t) = -4.9t^2 +C_2# where #C_2# is the height at #t = 0# (the initial height). SO #C = 1800# #m#

Thus #h(t) = 1800 - 4.9t^2#

#h(t) = 0# when #1800 - 4.9t^2 = 0# which happens at #t=sqrt(1800/4.9)#.

(We disregard the negative solution because our story involves only times after #t = 0#.)

Solving using Physics

#x_0=-1800m#
#v_0=0#
#a=-9.8ms^-2#

Function of displacement in term of time

Use the kinetic equation:
#x=v_0t+1/2at^2#

#x=0*t+1/2(-9.8)t^2#

#x=-4.9t^2#

Time taken for stone to hit the ground (travels 1800m down)

#-1800=-4.9t^2#

#t^2=1800/4.9#

#t=sqrt(1800/4.9#

#t~~19.1663s#