An enemy plane is flying horizontally at an altitude of 2km with a speed of `300ms^(-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 `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)` is `v_(x) = v_(0)costheta`
Let the shell hit the plane a point P and let t be the time taken for the shell to hit 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.
`uxxv = v_(x)xxt or u= v_(x) = v_(0) costheta`
` or costheta = (u)/(v_(0)) = (300)/(600)`

`=0.5 or theta = 60^(0)`.
Therefore, angle with the vertical = `90^(0)-theta=30^(0)`.