We know that the components of any vector along the x and y axes are given by:
![\begin{gathered} v_x=v\cos\theta \\ v_y=v\sin\theta \\ \text{ where:} \\ v\text{ is the magnitude of the vector.} \\ \theta\text{ is the angle from the }x\text{ axis to the vector. } \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/a5kzh8r2a9s7h5qljyb9fchhqot17pov70.png)
In this case we know the tha magnitude of the vector is 6.96 m/s; we also know that the angle from the x-axis to the vector is 51.5° (the sum of angles θ and the 26.5° angle); then we have:
![\begin{gathered} v_x=6.96\cos51.5=4.33 \\ v_y=6.96\sin51.5=5.45 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/kvezi1a3x266qt3bmzq0hmip9ofbh6fexi.png)
Therefore, the components of the vector are:
![\begin{gathered} v_x=4.33 \\ v_y=5.45 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/ctyyubp1hdt8xljilmtfknnn0o18ebnd6l.png)