Question

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

• A. $4s$
  
• B. $3s$
  
• C. $2s$
  
• D. $5s$
  

Answer 1

The correct option is C $2s$

Given : $u=u_x=10\sqrt{3}$ $m/s$ $u_y=0$

Acceleration in the x direction $a_x=-gsin{60}^o=-\frac{10\times \sqrt{3}}{2}=-8.66$ $m/s^2$

Acceleration in the y direction $a_y=-gcos{60}^o=-\frac{10}{2}=-5$ $m/s^2$

x direction :

Using $V_x=u_x+a_{x}t$

$\therefore$ $V_x=10\sqrt{3}-8.66t$

As the particle hits the inclined plane perpendicularly, thus it has final velocity in y direction only i.e $V_x=0$

$\therefore$ $0=10\sqrt{3}-8.66t$ $⟹t=2$ s

Hence the time of flight is $2$ s