Two inclined panes $OA$ and $OB$ having inclination ${30}^{°}$ and ${60}^{°}$ with the horizontal respectively intersect each other at $O$, as shown in figure. A particle is projected from point $P$ with a velocity $u=10\sqrt{3}m/s$ along a direction perpendicular to plane $OA$. If the particle strikes plane $OB$ perpendicular at $Q$

Which of the following is/are correct

Consider the motion of particle along the axes shown in figure. We have

$u_x=u, a_x=-g\sin 60°$

$u_y=0, a_y=-g\cos 60°$

(1) As the particle strikes the plane $OB$ perpendicularly,

$v_x=0\text{ as }{v}_x=u_x+a_{x}t$ or $0=u-g\sin 60°t\Rightarrow t=\frac{u}{g\sin 60°}=\frac{10\sqrt{3}}{10\times \frac{\sqrt{3}}{2}}=2s$

(2) Initial velocity along y-axis is zero. The velocity along y-axis after $2 s$; 

$v_y=u_y+a_{y}t$

$=0-g\cos 60°\times 2=-10\times \frac{1}{2}\times 2=-10 \mathrm{m}/\mathrm{s}$

(3) We have, $v_x^2=u_x^2+2a_{x}s$

Since $v_x=0$ and 

$a_x=g\sin 60°,u=10\sqrt{3} \mathrm{m}/\mathrm{s}$

$\therefore 0=(10\sqrt{3}{)}^2-2\times g\sin 60°\times (OQ)$ or $OQ=\frac{{10}^2\times 3}{2\times 10\times \frac{\sqrt{3}}{2}}=10\sqrt{3} \mathrm{m}$

Distance $PO=0+\frac{1}{2}g\sin 30°\times (2{)}^2$

$=\frac{1}{2}\times 10\times \frac{1}{2}\times 4=10 \mathrm{m}$

Therefore height $h$ of point $P$,

$h=PQ\sin 30°=10\times \frac{1}{2}=5 \mathrm{m}$

(4) Distance $PQ=\sqrt{P{O}^2+O{Q}^2}$

$=PQ=\sqrt{(10{)}^2+(10\sqrt{3}{)}^2}=20m$