1. **Problem statement:** Find the minimum initial velocity $v_0$ of a projectile launched from the origin so that it passes through the point $(3,4)$. 2. **Relevant formulas:** The projectile motion equations are: $$x = v_0 \cos(\theta) t$$ $$y = v_0 \sin(\theta) t - \frac{1}{2} g t^2$$ where $g = 9.8$ m/s² is the acceleration due to gravity. 3. **Goal:** Find the minimum $v_0$ such that the projectile passes through $(x,y) = (3,4)$ for some angle $\theta$ and time $t$. 4. **Express time $t$ from the $x$-equation:** $$t = \frac{3}{v_0 \cos(\theta)}$$ 5. **Substitute $t$ into the $y$-equation:** $$4 = v_0 \sin(\theta) \cdot \frac{3}{v_0 \cos(\theta)} - \frac{1}{2} g \left(\frac{3}{v_0 \cos(\theta)}\right)^2$$ Simplify: $$4 = 3 \tan(\theta) - \frac{1}{2} g \frac{9}{v_0^2 \cos^2(\theta)}$$ 6. **Rewrite using $\sec^2(\theta) = 1 + \tan^2(\theta)$:** $$4 = 3 \tan(\theta) - \frac{4.9 \cdot 9}{v_0^2} (1 + \tan^2(\theta))$$ 7. **Let $T = \tan(\theta)$, then:** $$4 = 3T - \frac{44.1}{v_0^2} (1 + T^2)$$ Rearranged: $$3T - 4 = \frac{44.1}{v_0^2} (1 + T^2)$$ 8. **Multiply both sides by $v_0^2$:** $$v_0^2 (3T - 4) = 44.1 (1 + T^2)$$ 9. **Solve for $v_0^2$:** $$v_0^2 = \frac{44.1 (1 + T^2)}{3T - 4}$$ 10. **To find minimum $v_0$, minimize $v_0^2$ with respect to $T$ where denominator $3T - 4 > 0 \Rightarrow T > \frac{4}{3}$.** 11. **Set derivative of $v_0^2$ with respect to $T$ to zero:** $$\frac{d}{dT} \left( \frac{44.1 (1 + T^2)}{3T - 4} \right) = 0$$ 12. **Derivative calculation:** $$\frac{44.1 \cdot (2T)(3T - 4) - 44.1 (1 + T^2) \cdot 3}{(3T - 4)^2} = 0$$ 13. **Simplify numerator:** $$44.1 [2T(3T - 4) - 3(1 + T^2)] = 0$$ $$2T(3T - 4) - 3(1 + T^2) = 0$$ $$6T^2 - 8T - 3 - 3T^2 = 0$$ $$3T^2 - 8T - 3 = 0$$ 14. **Solve quadratic:** $$T = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm 10}{6}$$ 15. **Possible $T$ values:** $$T_1 = 3, \quad T_2 = -\frac{1}{3}$$ Only $T_1=3$ satisfies $T > \frac{4}{3}$. 16. **Calculate minimum $v_0^2$ at $T=3$:** $$v_0^2 = \frac{44.1 (1 + 9)}{3 \cdot 3 - 4} = \frac{44.1 \times 10}{9 - 4} = \frac{441}{5} = 88.2$$ 17. **Minimum initial velocity:** $$v_0 = \sqrt{88.2} \approx 9.39$$ **Final answer:** The minimum initial velocity required is approximately $9.39$ m/s.