A particle is projected along an inclined plane as shown in figure$\left(\text{g}=\text{1}0{\text{m/s}}^{\text{2}}\right)$

According the motion of projectile on inclined plane

$\begin{array}{l}initial velocity along x axis (Incline plane)=u_x=u\cos (\alpha -\beta ); acceleration along x axis =a_x=-g\sin \beta ;\\ initial velocity along y axis{u}_y=u\sin (\alpha -\beta ); acceleration along y axis=a_y=-g\cos \beta \\ \alpha ={60}^{\circ };\\ \beta ={30}^{\circ };\\ \alpha -\beta ={60}^0-{30}^{\circ }={30}^{\circ }\\ u_x=10\cos 30=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}m/\sec , a_x=-10\sin 30=-5\mathrm{m}/\mathrm{s}\\ u_y=10\sin 30=10\times \frac{1}{2}=5\mathrm{m}/\sec a_y=10\cos 30=-10\times \frac{\sqrt{3}}{2}=-5\sqrt{3}\mathrm{m}/{\sec }^2\end{array}$

$\begin{array}{l}Time to reach from point 0 to point A is time of flight on the inclined plane\\ T=\frac{2u\sin (\alpha -\beta )}{g\cos \beta }\\ T=\frac{2\times 10\sin {30}^{\circ }}{10\times \cos {30}^{\circ }}\\ \\ T=\frac{2}{\sqrt{3}}\\ V_x\text{ at }A=u_x+a_{x}T=5\sqrt{3}-5\times \frac{2}{\sqrt{3}}=5\sqrt{3}-\frac{10}{\sqrt{3}}=\frac{5}{\sqrt{3}}m/\sec \\ V_y\text{ at }A=u_y+a_{y}T=5-5\sqrt{3}\times \frac{2}{\sqrt{3}}=-5\mathrm{m}/\sec \\ \mid \overrightarrow{V}{\mid }_{\text{ at }A}=\sqrt{V_x^2+V_y^2}=\sqrt{\frac{25}{3}+25}=\sqrt{\frac{100}{3}}=\frac{10}{\sqrt{3}}\mathrm{m}/\sec \end{array}$