1. **State the problem:** Danielle launches a ball from height $h$ with initial speed $v_0$ at angle $\theta$. A wall is at horizontal distance $L=25.0$ m with a hole at height $H$ where $H - h = \frac{1}{12}L$. Given $v_0^2 = \frac{3}{2} g L$, find the two angles $\theta_1$ (larger) and $\theta_2$ (smaller) so the ball passes through the hole. 2. **Known values and formulas:** - $L = 25.0$ m - $H - h = \frac{1}{12} L = \frac{25}{12} \approx 2.0833$ m - $v_0^2 = \frac{3}{2} g L$ - Gravity $g = 9.8$ m/s$^2$ Projectile motion equations: Horizontal position: $$x = v_0 \cos\theta \cdot t$$ Vertical position: $$y = h + v_0 \sin\theta \cdot t - \frac{1}{2} g t^2$$ 3. **Find $v_0$:** $$v_0 = \sqrt{\frac{3}{2} g L} = \sqrt{\frac{3}{2} \times 9.8 \times 25} = \sqrt{367.5} \approx 19.17 \text{ m/s}$$ 4. **Find time $t$ to reach the wall at $x=L$:** $$t = \frac{L}{v_0 \cos\theta}$$ 5. **Substitute $t$ into vertical position $y$ at $x=L$:** $$y = h + v_0 \sin\theta \cdot \frac{L}{v_0 \cos\theta} - \frac{1}{2} g \left(\frac{L}{v_0 \cos\theta}\right)^2$$ Simplify: $$y = h + L \tan\theta - \frac{g L^2}{2 v_0^2 \cos^2\theta}$$ 6. **Set $y = H$ and use $H - h = \frac{L}{12}$:** $$H = h + \frac{L}{12}$$ So, $$h + \frac{L}{12} = h + L \tan\theta - \frac{g L^2}{2 v_0^2 \cos^2\theta}$$ Subtract $h$ from both sides: $$\frac{L}{12} = L \tan\theta - \frac{g L^2}{2 v_0^2 \cos^2\theta}$$ Divide both sides by $L$: $$\frac{1}{12} = \tan\theta - \frac{g L}{2 v_0^2 \cos^2\theta}$$ 7. **Substitute $v_0^2 = \frac{3}{2} g L$ into the denominator:** $$\frac{g L}{2 v_0^2} = \frac{g L}{2 \times \frac{3}{2} g L} = \frac{g L}{3 g L} = \frac{1}{3}$$ So, $$\frac{1}{12} = \tan\theta - \frac{1}{3 \cos^2\theta}$$ 8. **Rewrite $\cos^2\theta$ in terms of $\tan\theta$:** Recall: $$\cos^2\theta = \frac{1}{1 + \tan^2\theta}$$ So, $$\frac{1}{3 \cos^2\theta} = \frac{1}{3} (1 + \tan^2\theta)$$ 9. **Substitute back:** $$\frac{1}{12} = \tan\theta - \frac{1}{3} (1 + \tan^2\theta)$$ Multiply both sides by 12 to clear denominators: $$1 = 12 \tan\theta - 4 (1 + \tan^2\theta)$$ Expand: $$1 = 12 \tan\theta - 4 - 4 \tan^2\theta$$ Bring all terms to one side: $$4 \tan^2\theta - 12 \tan\theta + 5 = 0$$ 10. **Solve quadratic equation for $u = \tan\theta$:** $$4 u^2 - 12 u + 5 = 0$$ Use quadratic formula: $$u = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 4 \times 5}}{2 \times 4} = \frac{12 \pm \sqrt{144 - 80}}{8} = \frac{12 \pm \sqrt{64}}{8} = \frac{12 \pm 8}{8}$$ Two solutions: $$u_1 = \frac{12 + 8}{8} = \frac{20}{8} = 2.5$$ $$u_2 = \frac{12 - 8}{8} = \frac{4}{8} = 0.5$$ 11. **Find angles $\theta$:** $$\theta_1 = \arctan(2.5) \approx 68.2^\circ$$ $$\theta_2 = \arctan(0.5) \approx 26.6^\circ$$ 12. **Final answer:** $$\boxed{\theta_1 = 68.2^\circ, \quad \theta_2 = 26.6^\circ}$$ These are the two angles ensuring the ball passes through the hole.