Question

A particle is projected up with a velocity of $v_0=10m/s$ at an angle of ${\theta }_0={60}^{\circ }$ with horizontal onto an inclined plane. The angle of inclination of the plane is ${30}^{\circ }$. The time of flight of the particle till it strikes the plane is
(take $g=10m/s^2$)

• A.
  $\frac{2}{2\sqrt{3}}\text{s}$
  
• B.
  $1/2\text{s}$
• C.
  $1\text{s}$
• D.
  $\frac{2}{\sqrt{3}}\text{s}$
  

Answer 1

The correct option is D

$\frac{2}{\sqrt{3}}\text{s}$

Step 1:Draw rough diagram for the given situation.

Step 2: Find time of flight.
Formula used: $T=\frac{2v_{\circ }\sin \theta }{g\cos \alpha }$

Given, projection velocity, $v_0=10m/s$

Angle of inclination, $\alpha ={30}^{\circ }$

Angle of projection from horizontal,${\theta }_0={60}^{\circ }$

So, $\theta =\theta -\alpha$
$\theta ={60}^{\circ }-{30}^{\circ }={30}^{\circ }$

Time of flight for oblique projectile motion,

$T=\frac{2v_{\circ }\sin \theta }{g\cos \alpha }$

$T=\frac{2\times 10\times \sin {30}^{\circ }}{10\times \cos {30}^{\circ }}$

$T=\frac{2}{\sqrt{3}}\text{s}$

Final answer: (C)