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A bomb traveling horizontally at $90 k m / h r$ suddenly explodes into two parts of mass ratio $1 : 3$ . Immediately after explosion, the lighter mass is found to be traveling vertically upwards with speed $360 \sqrt{3} k m / h r$ . The heavier one must be traveling at :

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$240 k m / h r$ downwards $30^{\circ}$ to horizontal (initial direction)

none

$240 k m / h r$ vertically downwards

$240 k m / h r$ downwards $60^{\circ}$ to horizontal (initial direction)

answer is C.

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Detailed Solution

$3 m V_{y} = m \times 360 \sqrt{3} V_{y} = 120 \sqrt{3} 4 m . 90 = 3 m V_{x} V_{x} = 120 \tan (θ) = \frac{V_{y}}{V_{x}} = \sqrt{3,} V = \sqrt{{V^{2}}_{x} + {v^{2}}_{y}}$

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