To solve this problem, we can break it down into two parts: the upward motion of the rock and its downward motion. Given the problem, we'll use the equations of motion under constant acceleration due to gravity.

Step 1: Analyze the motion and determine the initial speed

When you throw the rock upwards, it will rise, come to a stop at its peak, and then fall back down. Since the rock passes you again 6.00 seconds after it was thrown, we can use this time to determine its initial speed.

Let's denote:

• $v_0$ as the initial speed of the rock (which we need to find).
• $g$ as the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$).
• $t = 6.00 \, \text{s}$ as the total time the rock takes to return to the point from where it was thrown.

When the rock returns to the starting point after being thrown:

• The time taken to reach the highest point (where its velocity is zero) is $t_{\text{up}} = \frac{v_0}{g}$.
• The time taken to return back to the starting point from the highest point is the same, so the total time for the upward and downward journey to the starting point is $2t_{\text{up}} = 6.00 \, \text{s}$.

Thus, $t_{\text{up}} = \frac{6.00 \, \text{s}}{2} = 3.00 \, \text{s}$

Using $t_{\text{up}} = \frac{v_0}{g}$: $v_0 = g \cdot t_{\text{up}} = 9.81 \, \text{m/s}^2 \times 3.00 \, \text{s} = 29.43 \, \text{m/s}$

Step 2: Calculate the speed of the rock just before it reaches the water

Next, we need to find the speed of the rock just before it hits the water, which is 28.0 m below the point where it was thrown.

We can use the following kinematic equation: $v^2 = v_0^2 + 2g h$ Where:

• $v$ is the final speed just before hitting the water.
• $h = 28.0 \, \text{m}$ is the distance fallen.
• $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.
• $v_0 = -29.43 \, \text{m/s}$ (the negative sign indicates that the initial velocity is downward).

Substituting the values: $v^2 = (-29.43 \, \text{m/s})^2 + 2 \times 9.81 \, \text{m/s}^2 \times 28.0 \, \text{m}$

$v^2 = 866.3049 \, \text{m}^2/\text{s}^2 + 549.36 \, \text{m}^2/\text{s}^2$

$v^2 = 1415.6649 \, \text{m}^2/\text{s}^2$

$v = \sqrt{1415.6649 \, \text{m}^2/\text{s}^2} \approx 37.64 \, \text{m/s}$

Final Answer:

The speed of the rock just before it reaches the water is approximately 37.64 m/s.

Would you like further details on any part of the solution, or do you have any questions?

Here are 5 related questions to consider:

1. How would the presence of air resistance affect the rock's speed before hitting the water?
2. What would happen if the rock were thrown at an angle instead of straight up?
3. How can we determine the time it takes for the rock to hit the water after passing the point it was thrown from?
4. What would be the rock's velocity at its highest point?
5. How would the result change if the height from which the rock was thrown were different?

Tip: In free-fall problems, always pay attention to the direction of the initial velocity and the acceleration due to gravity to avoid sign errors in your calculations.