A fighter plane flying horizontally at an altitude of ` 1.5 km` with speed ` 720 kmh^(-1)` passes directly over head an anticraft gun.
At what angle from the vertical should the gun be fired from the shell with muzzle speed ` 600 ms^(-1)` to hit plane.
At what minimum altitude should the pilot fly the plane to avoid being hit ? ( Take g= 10 `ms^(-2)`).

Explanation

Consider that O is the position of gun and B is the position of fighter plane
Speed of shell `600 ms^(-1)`
Speed of plane `=720 kmh^(-1)=200 ms^(-1)`
Distance travelled by sheel `=(u sin theta) t=(600 sin theta)t`
Distance coverd by plane =ut
`=200t`
The shell will hit the plane if
`600 sin theta t =200 t`
`sin theta =(200)/(600)=(1)/(3)`
`theta=sin ^(-1)((1)/(3))=19.47^(@)` with the verticle .
To avoid the plane from being hit, the plane should fly at a height grater than the maximum height (H) attained by the shell .
`therefore H=(u^(2) sin^(2) (90-theta))/(2g)`
`=(u^(2) cos^(2) theta)/(2g)`
`=(600xx600 cos^(2)(19.47))/(2xx10)`
`=15998.4m`
`=16km`