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# A bomber plane moving at a horizontal speed of 20m/s releases a bomb at a height of 80m above ground as shown.At the same instant a hunter starts running from a point below it,to catch the bomb with speed 10m/s.After 2 seconds he realized that he cannot make it,he stops running and immediately holds 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 )

4 years ago
initial speed of bomb=20m/s
height=80 m
Therefore time of flight:-
80=0*t+1/2*g*t^2
t=4 s
distance travelled by hunter in 2s=10*2=20m
Range of bomb=20*4= 80m
Therefore distance left to cover is 60m
So V horizontal=60/2=30m/s
V vertical:-
Applying S=ut+1/2at^2
0=v*2-1/2*g*4
v=10m/s

Therefore resultant velocity=(Vx^2+Vy^2)^1/2
=( 30^2+10^2)^1/2
=(900+100)^1/2
=(1000)^1/2
=10(10)^1/2