Use a(t)= -9.8 meters per secondas the acceleration due to gravity.(Neglect air resistance)

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

To find the height of the rock as a function of time, we can use the following kinematic equation:

h(t) = h₀ + v₀t + (1/2)at²

In this equation:
- h(t) represents the height of the rock at time t
- h₀ is the initial height of the rock (the height of the rim)
- v₀ is the initial velocity of the rock (which is zero since the rock is dropped)
- a is the acceleration due to gravity (-9.8 m/s²)
- t is the time in seconds.

Since the rock is dropped, the initial velocity v₀ is zero. Therefore, the equation simplifies to:

h(t) = h₀ + (1/2)at²

Given that the rock is dropped from the rim of the Grand Canyon, the initial height h₀ is equal to the depth of the canyon, which is 1800 meters.

Plugging in the values, we have:

h(t) = 1800 + (1/2)(-9.8)t²

Now, to find the time it takes for the rock to hit the canyon floor, we need to find the value of t when the height h becomes zero.

0 = 1800 + (1/2)(-9.8)t²

To solve this equation, we can rearrange it to isolate t²:

(1/2)(-9.8)t² = -1800

Now, we can multiply both sides by 2 and divide by -9.8 to solve for t²:

t² = (-1800 * 2) / (-9.8)

t² = 367.35

Taking the square root of both sides, we find:

t ≈ 19.17 seconds

Therefore, it will take approximately 19.17 seconds for the rock to hit the canyon floor.

To find the height of the rock as a function of time, we can use the kinematic equation:

h(t) = h(0) + v(0)t + (1/2)at^2

Where:
h(t) = height of the rock at time t
h(0) = initial height (height of the rim)
v(0) = initial velocity (which is 0 as the rock is dropped)
a = acceleration due to gravity (-9.8 m/s^2)
t = time in seconds

In this case, since the rock is dropped from the rim, the initial height is the height of the rim, which is 1800 meters. The initial velocity is 0, as the rock is not given any initial velocity. Substituting these values into the equation, we get:

h(t) = 1800 + 0*t + (1/2)(-9.8)t^2

Simplifying further:

h(t) = 1800 - 4.9t^2

To find how long it will take for the rock to hit the canyon floor (height 0), we set h(t) = 0 and solve for t:

0 = 1800 - 4.9t^2

Rearranging the equation:

4.9t^2 = 1800

Dividing both sides by 4.9:

t^2 = 1800/4.9

t^2 ≈ 367.35

Taking the square root of both sides:

t ≈ √367.35

t ≈ 19.17 seconds

Therefore, it will take the rock approximately 19.17 seconds to hit the canyon floor at its deepest point.