A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km `h^(-1)` passes diretly overhead an anti-aircraft gun. At what angle from the vertical should the gun he fired for the shell with muzzle speed 600 m `s^(-1)` to hit the plane. (Take g = 10 m `s^(-1)`)

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.