Correct Answer - B::C
Velocity of man, `V_m = 3 km/hr`
BD= horizontal distance for resultant velocity R
X- component of resultant
`R= 5+3 cos theta`
`t= 0.5/(3 sin theta)`
Which is same of horizontal component of velocity.
H=BD
`=(5-3cos theta) (0.5/3 sin theta )`
`=(5+3 cos theta )/(6 sin theta)`
For H to be minimum `((dH)/(d theta))=0`
`gt d/(d theta) ((5+3 cos theta )/(6 sin theta ))=0`
`rarr 18 (sin^2 theta + cos^2 theta)-30 cos theta = 0`
`rarr 30 cos theta = 18`
`rarr cos theta = - 18/30 = - 3/5`
`sin theta = sqrt ( 1- cos^2 theta0=4/5`
`:. H= (5+3cos theta)/(6 sin theta)`
`= (5+3( 3/5))/)6xx4/5= (25-9)/24`
` = 16/24 = 2/3 km`.