To find the angle at which the football was thrown, we can use the equations of projectile motion. Given the horizontal distance ($d$) and the time of flight ($t$), we can first determine the horizontal and vertical components of the initial velocity.

1. Calculate the horizontal component of the initial velocity ($v_{0x}$): $v_{0x} = \frac{d}{t} = \frac{37.92 \, \text{m}}{2.06 \, \text{s}} \approx 18.38 \, \text{m/s}$

2. Assuming no air resistance, the vertical component of the velocity can be calculated using the formula for vertical motion: $d_y = v_{0y} \cdot t - \frac{1}{2} g t^2$ where $g \approx 9.81 \, \text{m/s}^2$ (acceleration due to gravity).

   However, since we don’t have the vertical distance covered, we can instead calculate the vertical component using the following equation derived from the tangent of the launch angle: $\tan(\theta) = \frac{v_{0y}}{v_{0x}}$

3. Using the vertical motion relation, we can express $v_{0y}$: $v_{0y} = g \cdot \frac{t}{2}$ This gives us the average vertical velocity during the time of flight. So: $v_{0y} = 9.81 \cdot \frac{2.06}{2} \approx 10.12 \, \text{m/s}$

4. Now, calculate the angle: $\tan(\theta) = \frac{v_{0y}}{v_{0x}} = \frac{10.12}{18.38}$ $\theta = \tan^{-1}\left(\frac{10.12}{18.38}\right) \approx 29.43^\circ$

Therefore, the angle at which the football was thrown is approximately 29.43 degrees.

Would you like more details on the calculations, or do you have any other questions? Here are some related questions you might consider:

1. How does air resistance affect projectile motion?
2. What is the maximum height the football would reach?
3. How would the angle change if the distance was greater?
4. Can you derive the equations of motion for projectile motion?
5. What factors influence the range of a projectile?

Tip: Remember that the angle of launch significantly affects the distance and height of a projectile.