A ball of mass 200 g rests on vertical post of height 20m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g=10m/s²):

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360 m/s

60 m/s

120 m/s

400 m/s

answer is C.

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Detailed Solution

$\begin{array}{l} t = \sqrt{\frac{2 h}{g}} = 2 \sec \\ V_{bullet} = 60 m/s \\ V_{ball} = 15 m/s \end{array}$
So using conservation of linear momentum $10 V = 200 \times 15 + 10 \times 60$
V=360 m/s

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