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 hit enemy plane when it is directly overhead and fires a shell with a muzzle speed of `600 ms^(-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 un and E that of the enemy plane flying horizontally with speed
`u = 300 ms^(-1)`, when the shell is fired with a speed `v_(0)` is `v_(x) = v_(0) cos theta`
Let the shell hit the plane a point P and let t be the time taken for the shell to hit the plane. It is clear that the shell will hit the plane, if the horizontal distance EP travelled by the plane in time t= the distance travelled by the shell in the horizontal direction in the same time,
i.e., `u xx v = v_(x) xx t` or `u = v_(x) = v_(0) cos theta`
or `cos theta = (u)/(v_(0)) = (300)/(600)`

= 0.5 or `theta = 60^(@)`.
Therefore, angle with the vertical `= 90^(@) - theta = 30^(@)`.