A block of mass m is placed on a surface having vertical cross...

At equilibrium, $tan\theta = \mu$.

Now, $tan\theta$ is slope of the curve. Therefore,

$tan\theta = \dfrac{dy}{dx} = \dfrac{x}{2}$

Now, 

$\dfrac{x}{2} = 0 . 5 ⇒ x = 1$

Required height 

$y = \dfrac{{x}^{2}}{4} = \dfrac{1}{4}m$.