We are tasked with finding the range of angles at which the rocks should be launched from the catapult to pass over the city wall. This is a projectile motion problem where the key objective is to find the angles that allow the heated rocks to cross the wall at 100 m away and a height of 15 m.

Step 1: Equation for projectile motion

The horizontal and vertical positions of a projectile launched with an initial speed $v_0$ at an angle $\theta$ are given by the following parametric equations:

$x(t) = v_0 \cdot \cos(\theta) \cdot t$ $y(t) = v_0 \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2$

Here:

• $v_0 = 80 \, \text{m/s}$ is the initial speed of the projectile,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $x = 100 \, \text{m}$ is the horizontal distance to the wall,
• $y = 15 \, \text{m}$ is the height of the wall.

We will use these equations to find the appropriate angles $\theta$ so that the projectile passes over the wall at $x = 100$ m and $y = 15$ m.

Step 2: Time to reach the wall

First, we find the time $t$ it takes for the projectile to travel horizontally to the wall 100 m away. From the horizontal position equation:

$x = v_0 \cdot \cos(\theta) \cdot t$

Solving for $t$:

$t = \frac{x}{v_0 \cdot \cos(\theta)} = \frac{100}{80 \cdot \cos(\theta)} = \frac{5}{4 \cdot \cos(\theta)}$

Step 3: Vertical motion equation

Now, we substitute this expression for $t$ into the vertical position equation:

$y = v_0 \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2$

Substitute $y = 15 \, \text{m}$, $v_0 = 80 \, \text{m/s}$, and $g = 9.8 \, \text{m/s}^2$:

$15 = 80 \cdot \sin(\theta) \cdot \frac{5}{4 \cdot \cos(\theta)} - \frac{1}{2} \cdot 9.8 \cdot \left( \frac{5}{4 \cdot \cos(\theta)} \right)^2$

Simplifying the first term:

$15 = 100 \cdot \tan(\theta) - \frac{1}{2} \cdot 9.8 \cdot \frac{25}{16 \cdot \cos^2(\theta)}$

$15 = 100 \cdot \tan(\theta) - \frac{122.5}{16 \cdot \cos^2(\theta)}$

$15 = 100 \cdot \tan(\theta) - \frac{7.65625}{\cos^2(\theta)}$

Step 4: Solve for $\theta$

This equation is transcendental and involves both $\tan(\theta)$ and $\cos(\theta)$. The best approach to solve this is numerically. We will find the solutions for $\theta$ within the range of $0^\circ$ to $90^\circ$.

Let me calculate the angles.It seems there was an issue finding distinct solutions in the first attempt. Let me refine the approach to find the correct angles within the expected range.The angle that allows the rocks to pass over the wall is approximately $12.9^\circ$.

Since only one angle solution is found in this case, the optimal angle to set the catapult is $12.9^\circ$.

If you have any more questions or need further details, feel free to ask!

Here are 5 related questions:

1. How does varying the initial speed of the projectile affect the required angle?
2. What happens to the angle if the wall height increases to 20 meters?
3. How would air resistance affect the calculation of the angle?
4. What is the maximum possible range of the projectile at this speed?
5. How would changing the gravitational constant (e.g., on another planet) affect the result?

Tip: In projectile motion, launching at complementary angles (like $45^\circ$ and $45^\circ$) can result in the same horizontal distance.