A bomber plane moving at a horizontal speed of $20 m/s$ releases a bomb at a height of $80 m$ above ground as shown. At the same instant a Hunter starts running from a point below it, to catch the bomb at $10 m/s$. After two seconds he realized that he cannot make it, he stops running and immediately hold his gun and fires in such direction so that just before bomb hits the ground, bullet will hit it. What should be the firing speed of bullet? (Take $g = 10 m/s^2$ )

In $2 sec,$ horizontal distance travelled by bomb $=20\times 2=40m$. In $2 sec.$ vertical distance travelled by bomb $=\frac{1}{2}\times 10\times 2^2 =20m$.

In $2 sec,$ horizontal distance travelled by Hunter $= 10\times 2 = 20 m$. Time remaining for bomb to hit ground $=\sqrt{\frac{2\times 80}{10}}-2 =2 sec$.

Let $V_x$ and $V_y$ be the velocity components of bullet along horizontal and vertical direction. Thus we use.

$\frac{2V_y}{g}=2\Rightarrow V_y=10 m/s$  

Considering the horizontal motion, The distance between bomb and bullet at t=2sec is 20m and relative speed is $\left(V_x-20\right)m/s$   

$\frac{20}{V_x-20}=2\Rightarrow V_x=30 m/s$

Thus velocity of firing is   $V=\sqrt{{V^2}_x+{V^2}_y}=10\sqrt{10} m/s$.