An enemy plane is flying horizontally at an altitude of `2km` with a speed of `300 ms^(-1)`.An army man with an anti-aircraft gun on the ground sights enemy plane when it is directly overhead and fires a shell with a muzzle speed of `600ms^(-1)`.At what angle with the vertical should the gun be fired so as to hit the plane?

Explanation

Let `G` be the position of the gun and `E` that of the enemy plane flying horizontally with speed.
`u=300ms^(-1)`, when the shell is fired with a speed `v_(0),v_(x)=V_(0)costheta`
The shell will hit the plane, if the horizontal distance `EP` travelled by the shell in time `t=` the distance travelled by the shell in the horizontal direction in the same time, i.e.
`uxxt=v_(x)xxt or u=v_(x) rArr u=v_(0)costheta`
or `costheta =u/v_(0)=300/600=0.5 or theta =60^(0)`
Therefore, angle with the vertical `=90^(@)-theta=30^(@)`.