velocity, the acceleration is zero. In this case:

$\mathrm{f}=\mathrm{mg} \sin \theta$ and $\mathrm{f}=\mu \mathrm{R}=\mu \mathrm{mg} \cos \theta$

$\therefore \quad \mu m g \cos \theta=m g \sin \theta$ or $\mu=\tan \theta \quad \ldots$ $(i)$

Case $II\, :$ The block is projected upward with initial velocity $u$ and hence it experiences downward acceleration a. In this case,

$\mathrm{mg} \sin \theta+\mu \mathrm{mg} \cos \theta=\mathrm{ma}$

or $\operatorname{mg} \sin \theta+\operatorname{mg} \tan \theta \cos \theta=\operatorname{ma}$

or $\operatorname{mg}(\sin \theta+\sin \theta)=\operatorname{ma}$ or $a=2 g \sin \theta \ldots$ $(ii)$

Let $x$ be the distance moved up the plane before the block comes to rest. Now,

${v^{2}-u^{2}=2 a s \text { or } 0-u^{2}=2(-2 g \sin \theta) x} $

$\therefore$      ${x=\frac{u^{2}}{4 g \sin \theta}}$