You drop a rock into a deep well and 4.1s later hear the splash.

How far down is the water? Neglect the travel time of the sound.
Express your answer using two significant figures.

To find the distance the rock traveled before hitting the water, we can use the equation of motion for free fall:

\(d = \frac{1}{2} g t^2\)

Where:
\(d\) is the distance,
\(g\) is the acceleration due to gravity (approximately 9.8 m/s^2), and
\(t\) is the time.

Substituting the given values into the equation:
\(d = \frac{1}{2} \times 9.8 \times (4.1)^2\)

Calculating:
\(d \approx 84\) meters

Therefore, the rock fell approximately 84 meters before hitting the water.

To determine how far down the water is in the well, we can use the equation of motion for freefall:

h = (1/2) gt^2

Where:
h is the height or depth of the well
g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
t is the time it takes for the rock to hit the water (4.1s in this case)

Plugging in the values, we have:

h = (1/2) * 9.8 m/s^2 * (4.1s)^2
h = (1/2) * 9.8 m/s^2 * 16.81s^2
h = 82.19 meters

Therefore, the water is approximately 82 meters deep in the well.

d = vt + at²/2

d = 0(4.1) + (-9.8)(4.1)²/2
d = -82 m

It is negative because this is the distance from ground level. In this case, the negative only gives its direction. Therefore, the answer is 82 m.