Answer
$v_x=v_0[1+(\frac{\pi}{L}c)^2\cos^2(\frac{\pi}{L}x)]^{-\frac{1}{2}}$
$v_y=\frac{v_0\pi c}{L}(\cos \frac{\pi}{L}x)[1+(\frac{\pi}{L}c)^2\cos^2(\frac{\pi}{L}x)]^{-\frac{1}{2}}$
Work Step by Step
$y=c\sin(\frac{\pi}{L}x)$
$\dot{y}=\frac{\pi}{L}c(\cos \frac{\pi}{L}x)\dot{x}$
$v_y=\frac{\pi}{L}cv_x(\cos \frac{\pi}{L}x)$
$v_0^2=v_y^2+v_x^2$
$v_0^2=v_x^2[1+(\frac{\pi}{L}c)^2\cos^2(\frac{\pi}{L}x)]$
$v_x=v_0[1+(\frac{\pi}{L}c)^2\cos^2(\frac{\pi}{L}x)]^{-\frac{1}{2}}$
$v_y=\frac{v_0\pi c}{L}(\cos \frac{\pi}{L}x)[1+(\frac{\pi}{L}c)^2\cos^2(\frac{\pi}{L}x)]^{-\frac{1}{2}}$
