Given data:
* The angle of the inclined plane is,
[tex]\theta=35^{\circ}[/tex]* The speed of the duck along the inclined plane is v = 5 m/s.
Solution:
The diagrammatic representation of the given case is,
From the triangle ABO,
[tex]\angle ABO=180^{\circ}-\angle AOB-\angle OAB[/tex]Substituting the known values,
[tex]\begin{gathered} \angle ABO=180^{\circ}-90^{\circ}-35^{\circ} \\ \angle ABO=55^{\circ} \end{gathered}[/tex]Thus, the inclined plane makes an angle of 55 degrees with the negative of the y-axis.
Similarly, the angle of the velocity of the duck along the inclined plane with the negative y-axis is 55 degrees.
Thus, the vertical component of the velocity is,
[tex]\begin{gathered} v_y=-v\cos (55^{\circ}) \\ v_y=-5\times\cos (55^{\circ}) \\ v_y=-2.88\text{ m/s} \end{gathered}[/tex]The horizontal component of the velocity is,
[tex]\begin{gathered} v_x=v\sin (55^{\circ}) \\ v_x=5\sin (55^{\circ}) \\ v_x=4.096\text{ m/s} \end{gathered}[/tex]Thus, the horizontal component of velocity is 4.096 meters per second, and the vertical component of velocity is -2.88 meters per second.
