A rock is dropped form the top of a cliff. It takes 12 seconds to hit the ground.

What is the height in feet of the cliff??

How fast is the rock moving when it hits the ground?

distance = |-16t^2| , here t is in seconds and distance is in feet

distance = 16(12^2) = 2304 feet

To find the height of the cliff, we can use the equation of motion. The equation that relates the height (h), initial velocity (u), time (t), and acceleration due to gravity (g) is:

h = u * t + 0.5 * g * t^2

In this case, the initial velocity (u) is zero since the rock is dropped. The acceleration due to gravity can be approximated as 32.2 feet per second squared (ft/s^2) near the surface of the Earth.

Substituting the given values into the equation, we have:

h = 0 * 12 + 0.5 * 32.2 * (12^2)

Simplifying the equation, we get:

h = 0 + 0.5 * 32.2 * 144

h = 0 + 16.1 * 144

h = 2318.4 feet

Therefore, the height of the cliff is approximately 2318.4 feet.

To determine the speed of the rock when it hits the ground, we can use the equation for velocity:

v = u + g * t

Since the rock starts from rest (u = 0), the equation becomes:

v = 0 + 32.2 * 12

Simplifying the equation, we find:

v = 386.4 ft/s

Therefore, the rock is moving at approximately 386.4 feet per second when it hits the ground.

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