Question

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 $\theta$ w.r.t the river flow. If AB =BC =400 m, what is the value of $\theta$?

• A. ${37}^{\circ }$
• B. ${53}^{\circ }$
• C. ${30}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B

${53}^{\circ }$

Resultant path of the swimmer is at ${45}^{\circ }$ with the bank. Therefore x and y-components of swimmer’s resultant velocity must be equal.
$tan{45}^{\circ }=\frac{v_y}{v_x}\Rightarrow v_y=v_x\Rightarrow v_R+v_{m}cos\theta =v_{m}sin\theta \Rightarrow 1+5cos\theta =5sin\theta$
Squaring both sides, $1+25co{s}^2\theta +10cos\theta =25-25co{s}^2\theta \Rightarrow 50co{s}^2\theta +10cos\theta -24=0$
Solving, we get $\theta ={53}^{\circ }$.