Speed of the bomber plane in horizontal direction u = 20m/s.
Height at which the bomb is released h =80m
If time of reaching ground be T then
1/2*g*T^2=80 => T =4s
During this time bomb will cover a distance = 20*4=80m
Hunter runs @10m/s and covers 10*2=20m in 2 sec. So the remaining horizontal distance (80-20)=60m is to be covered by the bullet fired by the hunter in remaining 2s..
Let the bullet be fired at speed v m/s.at an angle A with the horizontal to hit the bomb at the moment it reaches the ground.As the time of flight of the bullet is 2s we can write
vcosA*2=60=> vcosA= 30........(1)
And vsinA*2-1/2*10*2^2=0
=>vsinA=10.........(2)
From (1) and (2)
v^2=30^2+10^2
=>v = √1000=10√10m/s
