1. **State the problem:** We have an airplane flying horizontally at a height of 3000 ft with speed 400 ft/s. A package is released and follows the parabolic path given by $$y = 3000 - \frac{g}{2v^2} x^2$$ where $g = 32$ ft/s² and $v = 400$ ft/s. We want to find the "line of sight" angle $\theta$ between the horizontal and the line from the airplane to the target point $P$ on the water where the package lands. 2. **Understand the trajectory:** The package lands when $y=0$, so solve for $x$: $$0 = 3000 - \frac{32}{2 \times 400^2} x^2$$ 3. **Calculate the coefficient:** $$\frac{32}{2 \times 400^2} = \frac{32}{2 \times 160000} = \frac{32}{320000} = 0.0001$$ 4. **Find $x$ when $y=0$:** $$0 = 3000 - 0.0001 x^2 \implies 0.0001 x^2 = 3000 \implies x^2 = \frac{3000}{0.0001} = 30000000$$ $$x = \sqrt{30000000} = 5477.23 \text{ ft (approx)}$$ 5. **Find the slope of the trajectory at $x$:** The slope is the derivative $\frac{dy}{dx}$: $$\frac{dy}{dx} = - \frac{g}{v^2} x = - \frac{32}{400^2} x = - \frac{32}{160000} x = -0.0002 x$$ At $x = 5477.23$: $$\frac{dy}{dx} = -0.0002 \times 5477.23 = -1.0954$$ 6. **Find the angle of the tangent line to the trajectory at $P$:** The angle $\phi$ of the tangent line with the horizontal is: $$\phi = \arctan\left(\frac{dy}{dx}\right) = \arctan(-1.0954) = -47.5^\circ$$ 7. **Find the line of sight angle $\theta$:** The line of sight is the line from the airplane at $(0,3000)$ to the point $P$ at $(5477.23,0)$. Slope of line of sight: $$m = \frac{0 - 3000}{5477.23 - 0} = \frac{-3000}{5477.23} = -0.5477$$ Angle $\theta$ with horizontal: $$\theta = \arctan(-0.5477) = -28.7^\circ$$ Since angle below horizontal is positive in magnitude, $\theta = 29^\circ$ (nearest degree). **Answer:** The line of sight angle $\theta$ is approximately **29 degrees**. This means the pilot should sight the target at about 29 degrees below the horizontal to release the package so it lands at point $P$.