A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point ‘C’ starting from ‘A’. He swims with a speed of 5 km/hr at an angle θ w.r.t the river flow. If AB =BC =400 m, what is the value of θ?
53∘
Resultant path of the swimmer is at 45∘ with the bank. Therefore x and y-components of swimmer’s resultant velocity must be equal.
tan 45∘=vyvx⇒vy=vx⇒vR+vm cos θ=vm sin θ⇒1+5 cos θ=5 sin θ
Squaring both sides, 1+25cos2θ+10cosθ=25−25cos2θ⇒50 cos2 θ+10 cos θ−24=0
Solving, we get θ=53∘.