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
$\displaystyle \Psi^{I} (x,t)= A\, e^{-i\omega (t- \frac{1}{c}(x\,sin\theta -y\,cos\theta)) }$
the reflected potential is
$\displaystyle \Psi^{R} (x,t)= A\, e^{-i\omega (t- \frac{1}{c}(x\,sin\theta +y\,cos\theta)) }$

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

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

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