A launch plies between two points A and B on the opposite banks of a river always following the line AB. The distance S between points and B is 1200 m. The velocity of the river current `v= 1.9 m//s` is constant over the entire width of the river. The line AB makes an angle `alpha= 60^@` with the direction of the current. With what velocity u and at what angle beta to the line AB should the launch move to cover the distance AB and back in a time `t= 5 min`? The angle beta remains the same during the passage from A to B and from B to A.

Answer

`tan^-1 sqrt((3)/(17)) ; sqrt(85) ms^-1`

Explanation

From A to B : `AB = (v cos beta + u cos prop) t_1`
From B to A : `AB = (u cos beta - u cos prop)t_2`
`prop = 60^@, v sin beta = u sin prop`...(i)
`rArr 5 xx 60 = (1200)/(v cos beta + u cos prop) + (1200)/((v cos beta - v cos prop))`
Putting `u = sqrt(17), prop = 60^@`, we get
`v^2 cos^2 beta - 8v cos beta -(17)/(4) = 0 rArr v cos beta = (17)/(2)` ...(ii)
From Eqs. (i) and (ii), `tan beta = sqrt((3)/(17))`
`rArr beta = tan^-1 sqrt((3)/(17)) and v = (sqrt(85)) ms^-1`.