To solve this problem, we can break it down into two parts:

1. Time for the rock to return to your hand:

   • The total time for the rock to go up and come back down past you is given as $t = 9.00 \, \text{s}$.
   • This means the time for the rock to reach its maximum height and then return to your hand is $4.5 \, \text{s}$ (half of 9 seconds).

2. Calculating the initial velocity:

   • We'll use the kinematic equation to calculate the initial velocity: $v_f = v_0 - g t$ where:

     • $v_f = 0 \, \text{m/s}$ (the velocity at the maximum height is zero),
     • $v_0$ is the initial velocity (which we are solving for),
     • $g = 9.8 \, \text{m/s}^2$ (the acceleration due to gravity),
     • $t = 4.5 \, \text{s}$.

     Solving for $v_0$: $0 = v_0 - 9.8 \times 4.5$ $v_0 = 9.8 \times 4.5 = 44.1 \, \text{m/s}.$ So, the initial velocity of the rock is $44.1 \, \text{m/s}$.

3. Finding the velocity just before it reaches the water:

   • Now, we need to determine the velocity of the rock just before it hits the water 21 meters below the starting point.
   • We use the kinematic equation: $v^2 = v_0^2 + 2 g h$ where:
     • $v_0 = -44.1 \, \text{m/s}$ (the rock has reversed direction and is falling),
     • $g = 9.8 \, \text{m/s}^2$,
     • $h = 21.0 \, \text{m}$ (the distance to the water).

   Plugging in the values: $v^2 = (-44.1)^2 + 2 \times 9.8 \times 21$ $v^2 = 1944.81 + 411.6 = 2356.41$ $v = \sqrt{2356.41} \approx 48.55 \, \text{m/s}.$

Thus, the speed of the rock just before it reaches the water is approximately $48.55 \, \text{m/s}$.

---

Would you like more details on any of these steps or have any further questions?

Here are 5 related questions to consider:

1. How would the result change if we accounted for air resistance?
2. What is the maximum height the rock reaches above the bridge?
3. What is the total time taken for the rock to reach the water?
4. How would the problem change if the rock was thrown downward instead?
5. What if the initial velocity was twice as large?

Tip: Always split motion problems into parts—one for each phase of the movement, like upward and downward, to simplify calculations.