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A ball of mass 200 g rests on a vertical post of height 20m. A bullet of mass 10g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 20 m and the bullet at a distance 100 m from the foot of the post. The value of initial velocity (in m/s) of the bullet will be (if $g = 10 m / s^{2}$ ):

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answer is 250.

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Detailed Solution

$v_{1} = \frac{20}{\sqrt{\frac{2 h}{g}}}, v_{2} = \frac{100}{\sqrt{\frac{2 h}{g}}}$

$(0.01) u = (0.2) \frac{20 \sqrt{g}}{\sqrt{2 h}} + (0.02) \frac{100 \sqrt{g}}{\sqrt{2 h}}$

$u = 200 + 100 = 300 m s^{- 1}$

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