Question

The grand canyon is 1600 meters deep at its deepest point. A rock is dropped from the rim above this point. Express the height of the rock as a function of the time t in seconds. How long will it take the rock to hit the canyon floor? a(t) = -9.8m/sec/sec due to gravity.

Answer 1

I'm assuming you're not meant to derive the kinematics relationship \[\Delta x = \frac{ 1 }{ 2 }at^{2}\] so we're just gonna run with it.

if we call x the displacement from the rim to falling rock, and plug in our acceleration, the equation for x vs. t is
\[\ x = \frac{ -9.8 }{ 2 }t^{2}\]

To solve for the time to the bottom, let x = -1600
\[\ -1600 = \frac{ -9.8 }{ 2 }at^{2}\]
\[\ \sqrt \frac {-1600*2}{-9.8} = t\]
t = 18 seconds

Answer 2

thankyou very much.