Topic: Engineering Mechanics
As shown in the figure, the plane, having a mass of 8000 kg, is hoisted back until $theta = 60^circ$ and then the pull-back cable AC is released when the plane is at rest. Determine the speed of the plane just before it crashes into the ground at $theta = 30^circ$. (Neglect the size of the plane and effect of lift caused by the wings during the motion, assume $g=10 m/s²$)
- A. 23.98 m/s
- B. 12.1 m/s
- C. 9.56 m/s
- D. 6.76 m/s
Answer: B. 12.1 m/s
Explanation: By conservation of energy, the loss in potential energy equals the gain in kinetic energy. The vertical drop $Delta h = L cos(30^circ) - L cos(60^circ) = 20(sqrt{3}/2) - 20(1/2) = 17.32 - 10 = 7.32$ m. Equating energies: $mg Delta h = (1/2) m v²$. Thus, $v = sqrt{2g Delta h} = sqrt{2 * 10 * 7.32} = sqrt{146.4} = 12.1$ m/s.