As shown in the figure, the plane, having a mass of 8000 kg, is hoisted back until θ = 60$$\degree$$ 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 θ =30$$\degree$$ . (Neglect the size of the plane and effect of lift caused by the wings during the motion, assume g = 10 $$m/s^2$$​)

Correct option is B

​$$\textbf{Apply Conservation of Energy} \\\text{Let:} \\m = 8000\, \text{kg} \\g = 10\, \text{m/s}^2 \\\text{Radius of arc} = 20\, \text{m} \\\text{Initial angle} = \theta_1 = 60^\circ \\\text{Final angle} = \theta_2 = 30^\circ$$

$$\text{Use vertical height from point O:} \\h = R \cdot \cos(\theta) \\\\\text{Initial height: } h_1 = 20 \cos(60^\circ) = 20 \cdot 0.5 = 10\, \text{m} \\\text{Final height: } h_2 = 20 \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} \approx 20 \cdot 0.866 = 17.32\, \text{m} \\\\\text{Since it's falling from higher to lower:} \\\Delta h = h_1 - h_2 = 10 - 17.32 = -7.32\, \text{m} \\\\\text{Wait — this seems incorrect. Actually:} \\\text{At } \theta = 60^\circ, \text{ the vertical height is \textbf{higher} than at } \theta = 30^\circ \\\text{Correct heights are:} \\h_{60^\circ} = 20 \cdot \sin(60^\circ) = 20 \cdot 0.866 = 17.32\, \text{m} \\h_{30^\circ} = 20 \cdot \sin(30^\circ) = 20 \cdot 0.5 = 10\, \text{m} \\\\\text{So:} \\\Delta h = 17.32 - 10 = 7.32\, \text{m} \\\\\textbf{Step 3: Conservation of Mechanical Energy} \\\text{Initial PE = Final PE + Final KE} \\mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 7.32} = \sqrt{146.4} \approx \boxed{12.1\, \text{m/s}}$$​​​