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 anti-aircraft gun and E that of the enemy plane flying horizontally with a speed (V) = 300 m/s when a shell is fired with a speed u = 600 m/s in a direction 0 with the horizontal (in figure). The horizontal component of u is u cos 0.
Let the shell hit the plane at point T 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 ET traveled by the plane (moving with speed V) in time t is equal to the distance traveled in the horizontal direction by the shell (moving with speed u cos 0) in time t, i.e., if
`V xx t=u cos 0 xx t`
or `V= u cos 0`
`rarr cos 0 =(V)/(u) =(300 m//s)/(600 m//s) =(1)/(2) =0.5" or "0=60^(@)`
Therefore, angle with the vertical `=90^(@)-0 =30^(@)`
Thus, the shell should be fired in a direction `30^(@)` with the vertical.