hey im having trouble figuring this out. The question states that i am to find the second component of velocity?

there is a plane wave incident on a surface at (0,0).
the incident potential is
\Psi^{I} (x,t)= A\, e^{-i\omega (t- \frac{1}{c}(x\,sin\theta -y\,cos\theta)) }
the reflected potential is
\Psi^{R} (x,t)= A\, e^{-i\omega (t- \frac{1}{c}(x\,sin\theta +y\,cos\theta)) }

theta is the angle which the wave makes with the y axis.
i think there is an assumption that
\triangledown \times \mathbf{u} = 0 such that \mathbf{u}=\triangledown \Psi

First is the total velocity potential then just,
\Psi= \Psi^{I}+ \Psi^{R}

and then the second component of u would be
 \frac{\partial } {\partial y}(\Psi^{I}+\Psi^{R})