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Q. A ball of mass $200\, g$ rests on a vertical post of height $20\, m$. 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 $g=10\, m / s ^2$ ) :

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Solution:

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$ v _1=\frac{30}{\sqrt{\frac{2 h }{ g }}}, v _2=\frac{120}{\sqrt{\frac{2 h }{ g }}} $
$ (0.01) u =(0.2) \frac{30 \sqrt{ g }}{\sqrt{2 h }}+(0.01) \frac{120 \sqrt{ g }}{\sqrt{2 h }}$
$ u =300+60=360 \,ms ^{-1}$