This is a projectile motion problem where we need to determine the angle at which a ball was kicked, given the distance (range), initial speed, and horizontal ground level.

Given:

• Range $R = 30.0 \, \text{m}$
• Initial speed $v_0 = 8.0 \, \text{m/s}$
• Acceleration due to gravity $g = 9.81 \, \text{m/s}^2$

The formula for the range of a projectile in terms of its initial speed $v_0$ and angle $\theta$ is: $R = \frac{{v_0^2 \sin(2\theta)}}{g}$

Step-by-step solution:

1. Rearrange the formula to solve for $\theta$: $\sin(2\theta) = \frac{{R \cdot g}}{{v_0^2}}$

2. Substitute the given values: $\sin(2\theta) = \frac{{30.0 \cdot 9.81}}{{(8.0)^2}} = \frac{{294.3}}{{64}} \approx 4.598$

3. Since the sine function cannot exceed 1, this value is impossible. This suggests there is a mistake, potentially due to the input values. The range or speed seems incompatible for projectile motion under normal circumstances, given this angle equation.

Would you like to adjust the values, or clarify further? Let me know if you have questions.

Follow-up questions:

1. What happens to the angle when the initial speed increases while keeping the range constant?
2. Can you calculate the maximum range for this given initial speed?
3. What would the time of flight be for this projectile if the range and angle were known?
4. How would air resistance alter the calculated angle for this situation?
5. What is the significance of the angle 45° in projectile motion?

Tip:

The maximum range of a projectile is achieved when the launch angle is 45°, assuming no air resistance.