To solve this problem, we need to use the equations of projectile motion. Specifically, we are interested in finding the angles of launch that will allow the heated rocks to clear the city walls.

Step 1: Set up the coordinate system

We'll assume:

• The origin of the coordinate system is at the launch point (where the catapult is).
• The horizontal direction is $x$, and the vertical direction is $y$.

The rock is launched from a distance of 100 meters from the wall. The key parameters are:

• Initial speed $v_0 = 80 \, \text{m/s}$,
• Horizontal distance to the wall $d = 100 \, \text{m}$,
• Height of the wall $h = 15 \, \text{m}$,
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$.

Step 2: Parametric equations of projectile motion

The parametric equations for the horizontal and vertical positions of the rock are: $x(t) = v_0 \cos(\theta) t$ $y(t) = v_0 \sin(\theta) t - \frac{1}{2} g t^2$ where $t$ is the time of flight, and $\theta$ is the launch angle.

Step 3: Horizontal distance condition

We know that the rock must reach the wall when $x = 100 \, \text{m}$. From the horizontal motion equation, solve for the time $t$: $t = \frac{x}{v_0 \cos(\theta)} = \frac{100}{80 \cos(\theta)} = \frac{5}{4 \cos(\theta)}$

Step 4: Vertical height condition

At this time $t$, the rock must have a height of at least 15 meters to clear the wall. Substituting $t$ into the vertical motion equation: $y\left( \frac{5}{4 \cos(\theta)} \right) = v_0 \sin(\theta) \cdot \frac{5}{4 \cos(\theta)} - \frac{1}{2} g \left( \frac{5}{4 \cos(\theta)} \right)^2$ Simplifying this equation: $15 = \frac{80 \sin(\theta) \cdot 5}{4 \cdot 80 \cos(\theta)} - \frac{1}{2} \cdot 9.8 \cdot \left( \frac{5^2}{16 \cos^2(\theta)} \right)$ $15 = 5 \tan(\theta) - \frac{1}{2} \cdot 9.8 \cdot \frac{25}{16 \cos^2(\theta)}$ $15 = 5 \tan(\theta) - \frac{122.5}{16 \cos^2(\theta)}$

Step 5: Solve for $\theta$

We now have an equation involving $\theta$. Solving this equation will give us the range of angles for which the rocks will clear the wall. Let's solve this equation.The solutions we obtained include complex numbers, which are not physically meaningful for this problem (since the launch angles must be real). Let's try to solve this using a numerical approximation to find the real angles within a valid range.

We will attempt to numerically solve for the angles.It seems that there was a small issue with interpreting the value of $\frac{\pi}{2}$. I'll correct the code and proceed with solving for the valid angles.The valid range of launch angles for the catapult to successfully clear the city wall is approximately between $13.1^\circ$ and $85.5^\circ$.

Thus, the range of angles is:

$[13.1, 85.5] \, \text{degrees}$

Would you like more details or have any other questions?

Here are 5 related questions you might find interesting:

1. How can we adjust the angle if the initial speed of the catapult changes?
2. How would air resistance affect the range of angles?
3. What happens to the range of angles if the wall is taller?
4. How would the launch angle change if the catapult were placed further from the wall?
5. What if the city was located on a hill?

Tip: Always use degrees when dealing with projectile motion in real-world problems unless specifically asked to work in radians.