At anchored enemy ship is at a distance `180 sqrt(3) m` form the security cannon having a muzzle velocity of `60 m//s`.
(a) To what angle must the cannon be elevated to hit the ship.
(b) What is the time of flight.
(c ) How far should the ship be moved away from its initial position so that it becomes beyond the range of the cannon `(g = 10m//s^(2))`.

Explanation

For hitting the ship the range of cannon must be equalt ot he distance of ship from cannon, i.e,

`(u^(2)sin 2theta)/(g)=180sqrt(3)`
or `sin 2 theta=(180sqrt(3)xx10)/(60xx60)=sqrt(3)/(2)`
i.e., `2theta=60^(@)` or `120^(@)`
or `theta=30^(@)` or `60^(@)`
i.e., to hit the ship the cannon must be elevated at an angle of `30^(@)` or `60^(@)`
(b) As T=`(2u sin theta//g)`, depending on `theta` there are two times of flight.
`t_(1)=(2xx60)/(10)xxsin 30^(@)=6s`
and `t_(2)=(2xx60)/(10)xxsin 60^(@)=6sqrt(3)=10.4s`
(c) The maximum rang eof cannon (when `theta=45^(@)`)
`R_("max")=(u^(2))/(g)=(60xx60)/(10)=360`m
And as initially the ship is `180sqrt(3)` m, so to become out of maximum range of cannon, the ship shou ld be moved away from the harbour from its initial position by at least 360-`180sqrt(3)=48.6m`.