Respuesta :
Given information:
The velocity of the river is 2.8 m/s. The jet ski is at angle of 35° upstream. The velocity of the Jet ski relative to the ground is 9.5 m/s at an angle of 20° upstream.
The relative horizontal velocity of the Jet ski relative to the ground is,
[tex]\begin{gathered} \vec{v}_{jgx}=v_{jg}\cos (20\degree)(-\hat{i}) \\ =(9.5\text{ m/s})\times\cos (20\degree)(-\hat{i}) \\ \approx(8.927\text{ m/s})(-\hat{i}) \end{gathered}[/tex]The relative vertical velocity of the Jet ski relative to the ground is,
[tex]\begin{gathered} \vec{v_{jgy}}=v_{jg}\sin (20\degree)(\hat{j}) \\ =(9.5\text{ m/s})\times\sin (20\degree)(\hat{j}) \\ \approx(3.249\text{ m/s})(\hat{j}) \end{gathered}[/tex]The relative velocity of the jet ski relative to the ground is,
[tex]\begin{gathered} \vec{v}=\vec{v}+\vec{v} \\ =(8.927\text{ m/s})(-\hat{i})+(3.249\text{ m/s})(\hat{j}) \end{gathered}[/tex]The relative velocity of the river relative to the ground is,
[tex]\begin{gathered} \vec{v}_{rg}=v_{rg}(\hat{j}) \\ =(-2.8\text{ m/s})(\hat{j}) \end{gathered}[/tex]The relative velocity of the Jet ski relative to the river water is,
[tex]\begin{gathered} \vec{v}=\vec{v}-\vec{v} \\ =(8.3\text{ m/s})(-i)+(3.249\text{ m/s})(\hat{j})-(-2.8\text{ m/s})(\hat{j}) \\ =(8.927\text{ m/s})(-i)+(6.049\text{ m/s})(\hat{j}) \end{gathered}[/tex]The magnitude of the relative velocity of the Jet ski relative to the river water is,
[tex]\begin{gathered} v_{jr}=\sqrt[]{(8.927\text{ m/s})^2+(6.049\text{ m/s})^2} \\ \approx10.8\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the Jet ski relative to the water is 10.8 m/s.
