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?

2 Answers
Jul 20, 2017

h(t) = 1800 - 1/2g t^2 where g is the acceleration due to gravity. (See below.)

Explanation:

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.)

Jul 20, 2017

function of height of stone in respect to the starting point:
x=-4.9t^2

It takes approximately 19.1663s for the stone to hit the ground.

Explanation:

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