Question

Two inclined planes $OA$ and $OB$ having inclinations ${30}^{\circ }$ and ${60}^{\circ }$, respectively, intersect each other at $O$ as shown in the figure. A particle is projected from point $P$ with velocity $u=10\sqrt{3}\frac{m}{s}$ along a direction perpendicular to plane $OA$. If the particle strikes the plane $OB$ perpendicularly at $Q$, then calculate the maximum height (in $m$) from $O$ attained by the particle.

Answer 1

Step 1: Calculate $v$
Horizontal component of velocity $=$ Constant
$\Rightarrow u\cos {60}^{\circ }=v\cos {30}^{\circ }$
$\Rightarrow v=\frac{10\sqrt{3}/2}{\sqrt{3}/2}=10$
$\therefore v=10\frac{m}{s}$

Step 2: Calculate time to reach from $P$ to $Q$
Formula used:$v=u+at$
Applying $\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{at}(P\to Q)$
$\Rightarrow v\cos {30}^{\circ }\hat{i}-\sin {30}^{\circ }\hat{j}$
$=u\cos {60}^{\circ }\hat{i}+u\sin {30}^{\circ }-10t\hat{j}$
$\Rightarrow -\frac{10}{2}\hat{j}=10\sqrt{3}$
$\frac{\sqrt{3}}{2}\hat{j}-10t\hat{j}$
$\Rightarrow t=2s$

Step 3: Calculate displacement
Formula used: $s=\sqrt{x^2+y^2}$
$P\to Q:$
$\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$
$\overrightarrow{s}$
$=[10\sqrt{3}\cos {60}^{\circ }\hat{i}+10\sqrt{3}\cos {60}^{\circ }\hat{j}]2+\frac{1}{2}(-10\hat{j})2^2$
$\overrightarrow{s}=10\sqrt{3}\hat{i}+30\hat{j}-20\hat{j}=10\sqrt{3}\hat{i}+10\hat{j}$
$\overrightarrow{s}=10\sqrt{3}\hat{i}+10\hat{j}$
$\left|\overrightarrow{s}\right|=\sqrt{(10\sqrt{3}{)}^2+{10}^2}$
$\left|\overrightarrow{s}\right|=20m$
$\therefore PQ=20m$

Step 4: $h$ from trigonometry
in $\Delta OPQ$
$\frac{OP}{\sin {30}^{\circ }}=\frac{PQ}{\sin {90}^{\circ }}$
$OP=\frac{PQ}{2}=10m$
$OP=10m$
$\therefore h=OP\sin {30}^{\circ }$
$h=10\sin {30}^{\circ }=5m$
maximum height above $O$,
$H_{\text{max}}=h+\frac{(u\sin {60}^{\circ }{)}^2}{2g}=5+\frac{{15}^2}{20}=16.25m$

Final answer:$16.25m$