To solve this problem, we need to analyze the motion of the rock both horizontally and vertically using the principles of projectile motion. We'll break it down step-by-step.

Given:

• Initial velocity $v_0 = 25 \, \text{m/s}$
• Launch angle $\theta = 30^\circ$
• Height of the wall $h = 12 \, \text{m}$
• Acceleration due to gravity $g = 9.81 \, \text{m/s}^2$

Step 1: Break the velocity into components

We need to resolve the initial velocity into horizontal and vertical components:

• Horizontal component $v_{0x} = v_0 \cos(\theta)$
• Vertical component $v_{0y} = v_0 \sin(\theta)$

For $v_0 = 25 \, \text{m/s}$ and $\theta = 30^\circ$:

• $v_{0x} = 25 \cos(30^\circ) = 25 \times \frac{\sqrt{3}}{2} = 21.65 \, \text{m/s}$
• $v_{0y} = 25 \sin(30^\circ) = 25 \times \frac{1}{2} = 12.5 \, \text{m/s}$

Step 2: Determine the time the rock is in the air

To calculate the time of flight, we need to use the vertical motion equation:

$y = y_0 + v_{0y} t - \frac{1}{2} g t^2$

Where:

• $y_0 = 12 \, \text{m}$ is the initial height (the height of the castle wall),
• $y = 0$ is the final position (ground level),
• $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity,
• $v_{0y} = 12.5 \, \text{m/s}$ is the initial vertical velocity.

Substituting the values into the equation:

$0 = 12 + 12.5t - \frac{1}{2} \times 9.81 t^2$

This simplifies to:

$0 = 12 + 12.5t - 4.905t^2$

Rearranging:

$4.905t^2 - 12.5t - 12 = 0$

We can solve this quadratic equation using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 4.905$, $b = -12.5$, and $c = -12$. Substituting these values:

$t = \frac{-(-12.5) \pm \sqrt{(-12.5)^2 - 4(4.905)(-12)}}{2(4.905)}$ $t = \frac{12.5 \pm \sqrt{156.25 + 235.44}}{9.81}$ $t = \frac{12.5 \pm \sqrt{391.69}}{9.81}$ $t = \frac{12.5 \pm 19.8}{9.81}$

We get two possible values for $t$:

$t = \frac{12.5 + 19.8}{9.81} = \frac{32.3}{9.81} \approx 3.29 \, \text{seconds}$

We discard the negative root, as time cannot be negative.

Step 3: Calculate the horizontal distance

Now, to find the horizontal distance, we use the equation:

$d = v_{0x} \times t$

Substituting the values:

$d = 21.65 \, \text{m/s} \times 3.29 \, \text{s} \approx 71.3 \, \text{m}$

Final Answer:

The rock hits the ground approximately 71.3 meters from the castle wall.

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Would you like more details or a further breakdown of the steps? Here are some related questions you might find interesting:

1. How would the result change if the launch angle was increased to 45°?
2. How does the initial speed affect the range of the projectile?
3. What would happen if the castle wall was twice as high?
4. How can air resistance impact the horizontal distance traveled by the rock?
5. What is the maximum height reached by the rock during its flight?

Tip: Always remember to break down projectile motion into horizontal and vertical components, and use appropriate equations for each direction.