To solve this problem, let's break it down step-by-step using the equations of motion under gravity. We have the following given information:

• Height of the bridge, $h = 200$ m
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$
• Time delay between dropping the rock and throwing the baseball, $t_{\text{delay}} = 1.75$ s

Part (a): Initial Velocity of the Baseball

1. Time taken for the rock to hit the ground:

   • The rock is dropped from rest, so its initial velocity $u_{\text{rock}} = 0$.
   • Using the equation of motion:
     $h = \frac{1}{2} g t_{\text{rock}}^2$ Substituting the values: $200 = \frac{1}{2} \times 9.8 \times t_{\text{rock}}^2$ $200 = 4.9 \times t_{\text{rock}}^2$ $t_{\text{rock}}^2 = \frac{200}{4.9}$ $t_{\text{rock}} = \sqrt{\frac{200}{4.9}}$

2. Time for the rock: Let's calculate this.

3. Time taken for the baseball to reach the ground:

   • The baseball is thrown 1.75 seconds later, so its time to reach the ground: $t_{\text{baseball}} = t_{\text{rock}} - 1.75$

4. Equation for the baseball's motion:

   • The baseball is thrown with an initial velocity $u_{\text{baseball}}$ straight down.
   • Using the equation: $h = u_{\text{baseball}} t_{\text{baseball}} + \frac{1}{2} g t_{\text{baseball}}^2$ Substituting the known values, solve for $u_{\text{baseball}}$.

Let's do these calculations step-by-step.1. Time taken for the rock to hit the ground:
$t_{\text{rock}} \approx 6.39$ seconds

2. Time taken for the baseball to reach the ground:
   $t_{\text{baseball}} \approx 4.64$ seconds

3. Initial velocity of the baseball:
   $u_{\text{baseball}} \approx 20.38 \, \text{m/s}$

Part (b): Final Velocity of the Baseball

To find the final velocity of the baseball just before it hits the ground, we use the equation of motion: $v_{\text{baseball}} = u_{\text{baseball}} + g t_{\text{baseball}}$

Let's calculate this value.The final velocity of the baseball just before it hits the ground is approximately $65.84 \, \text{m/s}$.

Part (c): Final Velocity of the Rock

For the rock, which was dropped (initial velocity $u_{\text{rock}} = 0$), the final velocity is given by: $v_{\text{rock}} = g t_{\text{rock}}$

Let's calculate this value.The final velocity of the rock just before it hits the ground is approximately $62.61 \, \text{m/s}$.

Summary

• (a) Initial velocity of the baseball: $20.38 \, \text{m/s}$
• (b) Final velocity of the baseball: $65.84 \, \text{m/s}$
• (c) Final velocity of the rock: $62.61 \, \text{m/s}$

Do you want more details or have any questions about these calculations?

Related Questions

1. What would be the initial velocity of the baseball if the rock was dropped from a different height?
2. How would air resistance affect the velocities of both the baseball and the rock?
3. What if the baseball were thrown upwards instead of downwards?
4. How does the final velocity of the baseball change if thrown at an angle rather than straight down?
5. How would the scenario change if the delay was longer or shorter than 1.75 seconds?

Tip

Always ensure to keep track of time and velocity directions when dealing with projectile motion to avoid sign errors in your equations.