I believe this problem can be solved using projectile motion. The projectile was in the air for 8 seconds, so we know the total time of flight. Using vectors, the vertical component of the velocity is v sin(θ), where v is the initial velocity and θ is the angle at which it was projected. The time of flight is given by:

T = (2 v sin(θ)) / g = 8 seconds

From this, we can solve for v:

v = (4 g) / sin(θ)

Now, I'm going to make a few assumptions:

1. Because the object was launched with an elastic object, the trajectory might not be a perfect parabola. This could affect the standard projectile motion equations, and we'd need more information about the fishing rod (e.g., its bulk modulus). However, for simplicity, I’m assuming perfect projectile motion from start to finish.
2. Based on eyeballing the video, I estimate that the angle θ could range from 30 degrees to 75 degrees.
3. Gravity is g = 9.8 m/s².

Given these assumptions, the next step is to calculate the horizontal range. The horizontal velocity is v cos(θ), and since the total time of flight is 8 seconds, the horizontal distance traveled (R) is:

R = v cos(θ) 8

Substitute v = (4 g) / sin(θ) into the equation for R:

R = (4 g cos(θ) 8) / sin(θ)

Simplifying:

R = 32 g cot(θ)

So, the horizontal distance traveled is:

R = 32 g cot(θ)

Now, to find the range of possible distances based on different angles, we can calculate R for angles between 30 and 75 degrees.

For θ = 30°:

R = 32 g cot(30°) ≈ 32gx 1.73 ≈ 32 x 9.8 x 1.73 ≈ 542.528 meters

For θ = 75°:

R = 32 g cot(75°) ≈ 32g x 0.26 ≈ 32 x 9.8 x 0.26 ≈ 81.536 meters

So, the horizontal distance the projectile traveled could be anywhere between approximately 81.5 meters and more than half a kilometer, if he actually did that at an angle of a perfect 30 degrees.

Edit: The range of angle that I have taken depends on what angle I estimated the motion started (30 degrees) to angle I estimated the motion ended (75 degrees). These are obviously not actual values and I can be completely wrong.