A fighter plane flying horizontally at an altitude of `1.5 km ` with speed of `720 km h^9-10` passes directlu overhead an anticraft gun. At what anle fro the gun with muzzle speed ` 400 ms^(-1)` to hit the plane in shortest time ?`

Explanation

Let the figher plane be fluing borixontallu with a speed (u') at height `OP=1.5 km`. Let (P) be directly overhead an anticraft anticraft gun at (O). Let the shell be fired with velcity (u) making an angle `theta` with the vetical direction so that it hits the fighter place at (B) in shortest time, Fig.2 (d) .21.
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Resolving (u) into two rectangular cpmponents, we have,
`u cos theat`, vertically upwards and `u sin theta`, borizontally, If (t) is the time taken by the shell to hit the fighter plane, then horizontal distance travelled by the fighter plance in time (t) with velocity (u') is equal to the horizontal distance travelled by the shell in time (t) with velocity `u sin theta` i.e.,
`u' t= usin theta t or sin theta =u' //u`
Here, `u' =720 km h^(-1)`
`=720 xx (1000 m) xx (60 xx 60 s)^(-1)`
`=200 ms^(-1)`, and `u=400 ms^(-1)`
:. `sin theta =200//400 =1//2`
or `theta =30^(@)` with vertical`.