It seems like you're asking for the maximum horizontal distance of a shot or throw, but the problem appears to be incomplete. For a projectile (like a shot or a ball being thrown), the maximum horizontal distance is typically found using physics equations related to projectile motion.

The formula to calculate the maximum horizontal distance (also called the "range") for a projectile launched at an angle $\theta$ with an initial velocity $v_0$ is:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

Where:

• $R$ is the horizontal range (distance),
• $v_0$ is the initial velocity of the shot,
• $\theta$ is the launch angle,
• $g$ is the acceleration due to gravity ($32.2 \, \text{ft/s}^2$ on Earth).

For a more accurate solution, I would need values for the initial velocity and the launch angle. If you have this data, please provide it, and I can calculate the horizontal distance for you.

If you're asking for something else or the number "3030 feet" is part of the problem, feel free to clarify, and I'll assist further.

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Would you like to dive deeper into projectile motion? Here are some related questions for you:

1. How does the launch angle affect the range of a projectile?
2. What role does gravity play in projectile motion?
3. How would the initial speed of a shot affect the horizontal distance?
4. How do different air conditions (like wind or altitude) influence the maximum distance?
5. How can we calculate the time a projectile stays in the air?

Tip: If the launch angle is 45 degrees, the range is usually maximized for a given initial velocity.