Drone Package Release
A delivery drone flies horizontally at 12 m/s at a height of 45 m above a flat target. It releases a 2.0 kg package. The package has a small parachute that exerts a constant upward drag of 4.0 N after release; ignore air resistance before the parachute opens at 0.50 s. The target is a 2.0 m radius circle. The drone must release the package at a horizontal distance from the target center. Use g = 9.8 m/s². A safety rule requires the package to land within the circle and at an impact speed below 8 m/s. The drone can change speed by ±2 m/s or height by ±5 m.
1. Calculate the horizontal release distance ignoring drag and determine whether the no-drag drop would meet the impact-speed limit.
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Fall time from rest: t = sqrt(2h/g) = sqrt(90/9.8) = 3.03 s. Horizontal distance = 12 × 3.03 = 36.4 m. Vertical impact speed = 9.8 × 3.03 = 29.7 m/s. Total impact speed = sqrt(12² + 29.7²) = sqrt(144 + 882) = 32.0 m/s. This is far above 8 m/s, so the no-drag drop fails the speed rule even if the release distance were correct. The calculation also shows that horizontal speed contributes only a small part of the total impact speed; vertical fall dominates.
2. With the parachute, estimate whether the package can meet the 8 m/s impact-speed rule, stating assumptions.
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Assume the parachute opens at 0.50 s and drag is constant thereafter. Free fall for 0.50 s gives v = 4.9 m/s and fall distance = 1.23 m. Remaining height = 43.77 m. Net downward acceleration = (mg − 4.0)/m = (19.6 − 4.0)/2.0 = 7.8 m/s². Final vertical speed = sqrt(4.9² + 2 × 7.8 × 43.77) = sqrt(24.0 + 682.8) = 26.6 m/s. Adding horizontal velocity gives about 29.2 m/s. The package still fails the 8 m/s rule; a 4 N drag is much too small for a 2 kg package.
3. Recommend adjustments within the allowed speed and height changes, and justify whether they are sufficient.
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They are not sufficient. To land at 8 m/s from 45 m with constant acceleration, required net acceleration is a = v²/(2s) = 64/90 = 0.71 m/s². The needed drag is m(g − a) = 2.0(9.8 − 0.71) = 18.2 N, more than four times the current 4 N. Lowering height by 5 m to 40 m requires a = 64/80 = 0.80 m/s² and drag about 18.0 N, still far above 4 N. Changing horizontal speed by ±2 m/s barely affects vertical impact and may worsen total speed. The real fix is a larger parachute or guided descent, not the permitted speed and height trim.