An inclined plane is bent in such a way that the vertical...

At maximum ht. block will experience maximum friction force. Therefore if at this height slope of the tangent is $tan\theta,$ then $\theta =$Angle of repose

$∴ tan\theta = \dfrac{dy}{dx} = \dfrac{2x}{4} = \dfrac{x}{2} = 0 . 5$

$⇒ x = 1$ and therefore $y = \dfrac{{x}^{2}}{4} = 0 . 25m$

$= 25cm$

$∴$ Answer is $25cm$

(Assuming that $x$ & $y$ in the equation are given in meter)