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 after which the particle attains maximum height is:

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

Answer 1

The correct option is A

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

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

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

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

Time after which particle attains maximum height,
$t_{\text{ascent}}=\frac{u\left(\sin {\theta }_{\circ }\right)}{g}$
$t_{\text{ascent}}=\frac{10\times \left(\sin {60}^{\circ }\right)}{g}=\frac{\sqrt{3}}{2}\text{s}$

Final answer: (B)