A ball of mass 200g rests on a vertical post of height 20m . A...

Let initial velocity of bullet be $u$.

Given, mass of bullet $m = 10g$ and mass of ball $M = 200g$,

Initially ball is at height $5m$ and at rest, the only acceleration acting is due to gravity.

Time of flight of each ball and bullet is $t = \sqrt{\dfrac{2h}{g}}$.

Now, velocity of ball and bullet as ${v}_{1} = \dfrac{30}{\sqrt{\dfrac{2h}{g}}}and{v}_{2} = \dfrac{120}{\sqrt{\dfrac{2h}{g}}}$

As the collision is elastic, so applying conservation of linear momentum.

$mu + M\left(0\right) = M{v}_{1} + m{v}_{2}$

$⇒ \left(0 . 01\right)u = \left(0 . 2\right)\dfrac{30\sqrt{g}}{\sqrt{2h}} + \left(0 . 01\right)\dfrac{120\sqrt{g}}{\sqrt{2h}}$

$⇒ u = 300 + 60 = 360m{s}^{- 1}$