Two inclined planes OA and OB, with inclinations $30°$ and $60°$ with the horizontal respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity $u=10\sqrt{3} \mathrm{m}/\mathrm{s}$ along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicular at Q, calculate distance PQ. $(Take g=10 \mathrm{m}/{\mathrm{s}}^2)$

Let us choose the x and y directions along OB and OA respectively. Then,

$$\begin{gathered} u_x=u=10\sqrt{3} \mathrm{m}/\mathrm{s}, u_y=0 \\ {a}_x=-g\sin 60°=-5\sqrt{3} \mathrm{m}/{\mathrm{s}}^2 \\ {a}_y=-g\cos 60°=-5 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

At the point of collision at Q,  $v_x=0 \Rightarrow 0=u_x+a_{x}t=10\sqrt{3}-5\sqrt{3}t \Rightarrow t=2 \mathrm{s}.$.

Distance OQ = magnitude of displacement of particle along x-direction

$$\begin{gathered} s_x=u_{x}t+\frac{1}{2}{a}_x{t}^2=\left(10\sqrt{3}\right)\left(2\right)-\frac{1}{2}\left(5\sqrt{3}\right)(2{)}^2=10\sqrt{3} \mathrm{m} \\ \therefore \mathrm{OQ} = 10\sqrt{3} \mathrm{m}. \end{gathered}$$

Distance OP = magnitude of displacement of particle along y-direction

$$\begin{gathered} s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2=0+\frac{1}{2}\left(-5\right){\left(2\right)}^2=-10 \mathrm{m} \\ \therefore \mathrm{OP}=10 \mathrm{m}. \end{gathered}$$

$$\begin{gathered} PQ=\sqrt{(OP{)}^2+(OQ{)}^2}=\sqrt{(10{)}^2+(10\sqrt{3}{)}^2}=\sqrt{100+300}=\sqrt{400} \\ \text{or,} PQ=20 \mathrm{m} \end{gathered}$$