A block sliding along a horizontal frictionless surface with a velocity of 2 m s^-1 comes to the bottom of a frictionless inclined plane making an angle of 30° with the horizontal and comes to a stop after ascending the inclined plane as indicated. If g = 10 m s^-2, the value of ‘h’ is

Given an initial velocity of 2 m/s, a g-force of 10 m/s², and the $\theta ={30}^{°}$. Final velocity v = 0m/s because the block comes to a stop

The component along the block's motion direction is

$g\sin {30}^{°}=\frac{g}{2}$

by the equation of motion,

$v^2-u^2=2as$

$a=g\sin \theta =\frac{g}{2}$

$2^2-0^2=\frac{2gs}{2}$

$4=gs$

$s=\frac{h}{\sin \theta }$

$4=\frac{gh}{\sin \theta }$

$\sin {30}^{\circ }=\frac{1}{2}$

$h=\frac{4}{2g} = 0.2m$

Hence the correct answer is 0.2m.