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A projectile is fired at a speed of $100 m s^{- 1}$ at an angle of $37^{0}$ above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the smaller coming to rest. Then the distance of heavier part from the launching point is

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480 m

960 m

1120 m

1440 m

answer is C.

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Detailed Solution

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

$R = \frac{\frac{m}{4} \times \frac{R}{2} + \frac{3 m}{4} x_{2}}{\frac{m}{4} + \frac{3 m}{4}}$

$R = \frac{R}{8} + \frac{3 x_{2}}{4}$

$R - \frac{R}{8} = \frac{3 x_{2}}{4}$

$x_{2} = \frac{7 R}{6} = \frac{7}{6} \frac{U^{2} \sin 2 θ}{g}$

$x_{2} = \frac{7}{6} U^{2} \times \frac{2 \sin θ \cos θ}{g}$

$x_{2} = \frac{7}{6} \times 100 \times 100 \times 2 \times \frac{3}{5} \times \frac{4}{5} \times \frac{1}{10}$

$= 7 \times 160$

$x_{2} = 1120 m$

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