explain---**Question Stem:** A plane is moving horizontally through the sky with speed v. A package is dropped from the plane and travels a horizontal distance x from the point of release to where it lands on a target. If air resistance is negligible, the height from which the package is released is **Options:** (A) $\frac{1}{2} g \sqrt{\frac{x}{v}}$ (B) $\frac{1}{2} gt^2$ (C) $\frac{1}{2} g \left(\frac{x}{2v}\right)$ (D) $\frac{1}{2} g \left(\frac{x}{v}\right)^2$ (E) $\frac{1}{2} g \left(\frac{v}{x}\right)^2$ **Chart/Diagram Description:** * **Type:** Trajectory path / Projectile motion diagram. * **Main Elements:** * A horizontal line representing the ground or landing surface. * A curved line representing the path of the dropped object (package), starting from a point above the horizontal line and ending on the horizontal line. * An arrow on the curve indicates the direction of motion along the path. * A small circle symbol at the starting point of the trajectory. * A point labeled 'P' is located on the curved path. * The curve is a downward-opening arc, consistent with a parabolic trajectory under constant gravitational acceleration.

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Let's analyze this projectile motion problem. A plane moving horizontally with speed v drops a package. The package travels a horizontal distance x before hitting the target. We need to find the height h from which it was released. This is a classic physics problem involving two-dimensional motion under gravity. Now let's analyze the horizontal motion. The package maintains the plane's horizontal velocity v throughout its flight. Since there's no horizontal acceleration, the horizontal distance x equals velocity times time. This gives us the equation x equals v times t. Solving for time, we get t equals x divided by v. This time of flight is crucial for our vertical motion analysis. For the vertical motion, the package starts with zero vertical velocity since it's dropped, not thrown. It accelerates downward due to gravity g. Using the kinematic equation for constant acceleration, the height h equals one-half g times t squared. This parabolic relationship shows how the vertical distance increases quadratically with time. Now let's combine our equations to find the height. From horizontal motion, we have x equals v times t, so t equals x over v. From vertical motion, we have h equals one-half g t squared. Substituting our expression for t into the height equation, we get h equals one-half g times the quantity x over v squared. This matches option D perfectly. Let's review all the options. Option A has a square root which doesn't match our derivation. Option B is incomplete as it doesn't substitute for time. Option C has the wrong coefficient. Option E has velocity and distance inverted. Option D is correct: h equals one-half g times x over v squared. This makes physical sense - greater horizontal distance means longer flight time and greater height, while faster plane speed means shorter flight time and lower release height.