A bullet of mass 50g is fired from below into the bob of mass 450g of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of 1.8 m. Find the speed of the bullet. Take g =10 m / s²

Let the speed of the bullet be v. Let the common velocity of the bullet and the bob, after the bullet is embedded into the bob, is V. By the principle of conservation of the linear momentum  
$\mathrm{V}=\frac{(0.05\mathrm{kg})\mathrm{v}}{0.45\mathrm{kg}+0.05\mathrm{kg}}=\frac{\mathrm{v}}{10}$
The string becomes loose and the bob will go up with a deceleration of $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2.$
As it comes to rest at a height of 1.8m, using the equation ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{ax}$
$1.8\mathrm{m}=\frac{(\mathrm{v}/10{)}^2}{2\times 10\mathrm{m}/{\mathrm{s}}^2}\text{ or }\mathrm{v}=60\mathrm{m}/\mathrm{s}$