Question

From the previous problem at what angle to the horizontal must be steered in order to reach a point on the opposite bank directly east from the starting point? (Assuming his speed w.r.t. to the river is 4 m/s)

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B

${30}^{\circ }$

We know that ${\overrightarrow{v}}_{man}$ should be horizontal let us assume that he steers the boat at some angle to the

horizontal such that, his velocity w.r.t ground is horizontal.

This means, his velocity w.r.t to the river will make angle $\theta$ with the x axis.

We know his speed w.r.t to the river is 4 m/s.

$\therefore {\overrightarrow{v}}_m={\overrightarrow{v}}_{mr}+{\overrightarrow{v}}_r$

= $4cos\theta \hat{i}+4sin\theta \hat{j}-2\hat{j}$

= $4cos\theta \hat{i}+(4sin\theta -2)\hat{j}m/s$

To reach a point exactly opposite his starting point,his vertical must be zero.

$\therefore 4sin\theta -2=0$

$\Rightarrow sin\theta =\frac{1}{2}=\frac{1}{2}$

$\Rightarrow \theta ={30}^{\circ }$

$\therefore$ He must start at ${30}^{\circ }$ to the horizontal

Method - 2

Using triangle law, we knew

${\overrightarrow{v}}_{mr}+{\overrightarrow{v}}_r={\overrightarrow{v}}_m$

And ${\overrightarrow{v}}_m$ is horizontal

$\therefore sin\theta =\frac{2}{4}=\frac{1}{2}\therefore \theta ={30}^{\circ }$