1. **State the problem:** A bullet is shot at an angle of 25 degrees above the ground and hits a target 301.5 meters away. We need to find the maximum height the bullet reaches. 2. **Relevant formulas:** For projectile motion, the horizontal range $R$ and maximum height $H$ are given by: $$R = \frac{v_0^2 \sin(2\theta)}{g}$$ $$H = \frac{v_0^2 \sin^2(\theta)}{2g}$$ where $v_0$ is the initial velocity, $\theta$ is the launch angle, and $g = 9.8$ m/s$^2$ is the acceleration due to gravity. 3. **Find initial velocity $v_0$ using the range formula:** $$301.5 = \frac{v_0^2 \sin(2 \times 25^\circ)}{9.8}$$ Calculate $\sin(50^\circ)$: $$\sin(50^\circ) \approx 0.7660$$ So, $$301.5 = \frac{v_0^2 \times 0.7660}{9.8}$$ Multiply both sides by 9.8: $$301.5 \times 9.8 = v_0^2 \times 0.7660$$ $$2954.7 = v_0^2 \times 0.7660$$ Divide both sides by 0.7660: $$v_0^2 = \frac{2954.7}{0.7660}$$ $$v_0^2 \approx 3857.7$$ Take the square root: $$v_0 = \sqrt{3857.7} \approx 62.1 \text{ m/s}$$ 4. **Calculate maximum height $H$:** $$H = \frac{v_0^2 \sin^2(25^\circ)}{2 \times 9.8}$$ Calculate $\sin(25^\circ)$: $$\sin(25^\circ) \approx 0.4226$$ Square it: $$\sin^2(25^\circ) = (0.4226)^2 = 0.1786$$ Substitute values: $$H = \frac{3857.7 \times 0.1786}{19.6}$$ $$H = \frac{688.9}{19.6} \approx 35.1 \text{ meters}$$ **Final answer:** The maximum height reached by the bullet is approximately $35.1$ meters.