1. **State the problem:** A long jump athlete leaves the ground at an angle of 30 degrees and lands 8.9 meters away. We need to find the initial velocity $v_0$ at which he left the ground. 2. **Relevant formula:** The horizontal range $R$ of a projectile launched at an angle $\theta$ with initial velocity $v_0$ is given by: $$ R = \frac{v_0^2 \sin(2\theta)}{g} $$ where $g = 9.8$ m/s² is the acceleration due to gravity. 3. **Given values:** - $R = 8.9$ m - $\theta = 30^\circ$ - $g = 9.8$ m/s² 4. **Rearrange the formula to solve for $v_0$:** $$ v_0^2 = \frac{R g}{\sin(2\theta)} $$ 5. **Calculate $\sin(2\theta)$:** $$ \sin(2 \times 30^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 $$ 6. **Substitute values:** $$ v_0^2 = \frac{8.9 \times 9.8}{0.866} $$ 7. **Calculate numerator and denominator:** $$ 8.9 \times 9.8 = 87.22 $$ 8. **Calculate $v_0^2$:** $$ v_0^2 = \frac{87.22}{0.866} \approx 100.72 $$ 9. **Take the square root to find $v_0$:** $$ v_0 = \sqrt{100.72} \approx 10.04 \text{ m/s} $$ **Final answer:** The athlete left the ground with an initial velocity of approximately **10.04 m/s**.