Answer:5.99 m/s
Explanation:
Given
Speed of river [tex](v_r)[/tex]=4.1 m/s
Width of river =260 m
Speed of boat[tex](v_{br})[/tex]=7.5 m/s
launch angle[tex]=37^{\circ}[/tex]
[tex]v_{br}=7.5(cos37\hat{j}-sin37\hat{i})[/tex]
[tex]v_r=4.1\hat{i}[/tex]
Velocity of boat with respect to ground
[tex]v_b=v_{br}+v_r[/tex]
[tex]v_b=7.5(cos37\hat{j}-sin37\hat{i})+4.1\hat{i}[/tex]
[tex]v_b=(-7.5sin37+4.1)\hat{i}+7.5\times cos37\hat{j}[/tex]
[tex]v_b=-0.413\hat{i}+5.98\hat{j}[/tex]
[tex]|v_b|=\sqrt{0.413^2+5.98^2}[/tex]
[tex]|v_b|=5.99 m/s[/tex]
Direction
[tex]tan\theta =\frac{5.98}{0.413}=14.479[/tex]
[tex]\theta =86.049^{\circ}[/tex] north of west
