Question

Two inclined planes $OA$ and $OB$ having inclination (with horizontal) ${30}^o$ and ${60}^o$, respectively, intersect each other at $O$ as shown in Figure. A particle is projected from point $P$ with velocity $u=10\sqrt{3}m{s}^{-1}$ along a direction perpendicular to plane $OA$. If the particle strikes plane $OB$ perpendicularly at $Q$, calculate the
a. velocity with which particle strikes the plane $OB$.
b. time of flight.
c. vertical height h of P from O.
d. maximum height from O, attained by the particle.
e. distance PQ.

Answer 1

Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane OA. It means is is parallel to plane OB.

At the instant of collision of the particle with OB, its velocity is perpendicular to OB or the velocity component parallel to OB is zero.

For considering the motion of particle parallel to plane OB,

$\overrightarrow{u}=10\sqrt{3}0m{s}^{-1}$

Acceleration $=-gsin{60}^o=-5\sqrt{3}m{s}^{-2}$

$u=0,t=?,S=?$

Using $v=u+at,t=2s$

$s=ut+\frac{1}{2}a{t}^2$ or $OQ=10\sqrt{3}m$

Now considering the motion of the particle normal to plane OB. Initial velocity = 0, acceleration $=gcos{60}^o=5m{s}^{-2}$

$T=2s,v=?,s=PO=?$

Using $v=u+at,v=10m{s}^{-1}$

$s=ut+\frac{1}{2}a{t}^{2}orOP=10m$

$h=POsin{30}^o=10\times sin{30}^o=5m$

The inclination of $\overrightarrow{u}$ with the vertical is ${30}^o$. therefore, its vertical component is $ucos{30}^o=15m{s}^{-1}$ (upward).

Considering vertically upward motion of the particle from P, initial velocity = 15 m s${}^{-1}$,acceleration = g -10 m s${}^{-2}$, $v=0.S=H=$?

Using, $v^2=u^2+2as,H=11.25m$

Therefore, maximum height reached by the particle above $O=h+H=16.25m$

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