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 of negligible height starts running from a point below it, to catch the bomb with speed `10 m//s`. After two seconds he relized 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)`)

Options

A. `10 m//s`
B. `20sqrt10 m//s`
C. `10sqrt10 m//s`
D. None of these

Answer

C

Explanation

In `2 sec.` horizontal distance travelled by bomb `=20xx2=40m`.In `2 sec.` vertical distance travelled by bomb`=1/2xx10xx2^(2)=20m`.
In `2 sec.` horizontal distance travelled by Hunter `=10xx2=20m`.
Time remaining for bomb to hit ground
`=sqrt((2xx80)/10)-2=2 sec`.
Let `V_(x) and V_(y)` be the velocity components of bullet along horizontal and vertical direction. Thus we use.
`(2V_(y))/g=2rArr V_(y)=10 m//s` and
`20/(V_(x)-20)=2rArr V_(x)=30 m//s`
Thus velocity of firing is
`V=sqrt(V_(x)^(2)+V_(y)^(2))=10sqrt10 m//s.`
C) projectile motion on inclined plane