At aharbour enemy ship is at a distance `180sqrt(3)m` form the security cannot having a muzzle velocity of `60m//s`.
(a) To what angle must the cannot be elevated to hit the ship.
(b) What is the time of flight.
(c ) How for should the ship be moved away from its initial position so that it becomes beyond the range of the cannot `(g = 10m//s^(2))`.

Explanation

(a) For hitting the ship the range of cannon must be equal to the distance of ship from cannot i.e.,
Range `= 180sqrt(3)`

`(u^(2)sin 2theta)/(g) = 180 sqrt(3)`
r `sin 2 theta = (180sqrt(3)xx10)/(60xx60) = (sqrt(3))/(2)`
`{:(i.e.,2theta=60^(@),or,,120^(@)),(or,theta=30^(@),or,,60^(@)):}`
(a) The cannot mist be elevated at an angle of `30^(@)` or `60^(@)`
(b) As `T = (2usin theta//g)`, depending on `theta` there are two times of flight which are given by
`t_(1) = (2xx60)/(10) xx sin30 = 6s`
and `t_(2) = (2xx60)/(10) xx sin60 = 6sqrt(3) = 104s`.
(c) The maximum range of cannot (where `theta =45^(@))`
`R_(max) = (u^(2))/(g) = (60xx60)/(10) = 360m`
And as initially the ship is `180sqrt(3)m`, so to become out of maximum range of cannot, the ship should be moved away form the harbour form its initial position by at least
`360 - 180sqrt(3) = 48.6m.`