A fighter plane flying horizontally at an altitude of ` 1.5 km` with speed ` 720 kmh^(-1)` passes directly over head an anti-aircraft 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

Velocity of plane, `V_(p) = 720 xx 5//180ms^(-1)=200ms^(-1)`
Velocity of shell =`600ms^(-1)`
`sintheta=200/600 = 1/3`
or `theta=sin^(-1)(1/3)=19.47^(@)`.
This angle is with the vertical
Let h be the required minium height.
Using equation
`v^(2)-u^(2)=2as`, we get
`(0)^(2) - (600costheta)^(2)=-2 xx 10xx h`
or `h=(600 xx 600(1-sin^(2)theta))/(20)`
`= 30 xx 600(1-1/9)= 8/9 xx 30 xx 600m =16 km.