You throw a small rock straight up from the edge of a high way bridge that crosses a river. The rock passes you on it's way down, 6.00s after it was thrown. What is the speed of the rock just before it reaches the water 28.0m below the point where the rock left your hand? Ignore air resistance

To solve this problem, we can use the kinematic equations of motion under uniform acceleration due to gravity. Since air resistance is ignored, the acceleration \( a \) is constant and equal to \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 1: Understanding the Problem

- A rock is thrown straight up and passes back by the thrower 6.00 seconds after it was thrown.
- We need to find the speed of the rock just before it hits the water, which is 28.0 meters below the point of release.

### Step 2: Define Variables

Let's define the following:

- \( u \): Initial velocity of the rock (upwards).
- \( t \): Time taken to pass back the thrower and return to the same level, \( t = 6.00 \) seconds.
- \( g \): Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \).
- \( v \): Final velocity just before hitting the water.
- \( h \): Height from the thrower’s hand to the water, \( h = 28.0 \, \text{m} \).

### Step 3: Find the Initial Velocity

The total time \( t \) is the sum of the time taken to reach the highest point and the time to return to the initial level. Since these times are equal:

\[
t_{\text{up}} = t_{\text{down}} = \frac{t}{2} = 3.00 \, \text{s}
\]

Using the equation for the time to reach the highest point:

\[
t_{\text{up}} = \frac{u}{g}
\]

\[
3.00 = \frac{u}{9.8}
\]

\[
u = 3.00 \times 9.8 = 29.4 \, \text{m/s}
\]

### Step 4: Calculate the Final Velocity

We want the velocity \( v \) of the rock just before it reaches the water. We'll use the kinematic equation that relates velocity, initial velocity, acceleration, and displacement:

\[
v^2 = u^2 + 2gh
\]

Here \( h = 28.0 \) m (downward, so \( h \) is positive in this context):

\[
v^2 = 0^2 + 2 \times 9.8 \times 28.0
\]

\[
v^2 = 548.8
\]

\[
v = \sqrt{548.8}
\]

\[
v \approx 23.42 \, \text{m/s}
\]

### Conclusion

The speed of the rock just before it reaches the water is approximately **23.42 m/s**.

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Do you want more details or have any questions? Here are some related questions:

1. What would be the speed of the rock if it fell from a greater height?
2. How would air resistance affect the final speed of the rock?
3. What is the maximum height reached by the rock above the bridge?
4. How would the time of flight change if the initial velocity was doubled?
5. If the rock was thrown horizontally instead of vertically, what would be its speed before hitting the water?

**Tip:** When solving projectile motion problems, always separate the motion into vertical and horizontal components for clarity.

Learn how to calculate the speed of a rock just before it hits the water after being thrown vertically upwards. This physics problem involves using kinematic equations of motion under gravity, ignoring air resistance. Find the final velocity and understand the motion dynamics. Suitable for high school physics students.

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