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:

Answer 1

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)