A block slides down an inclined plane of slope of angle q with a constant velocity v' It is then projected up the plane with an initial velocity u. the distance upto which it will rise before coming to rest is

The different forces in the two ca6es are shown in fig.

Case (i). As the block slides down with constant velocity, the acceleration is zero. In this case

$\mathrm{f}=\mathrm{mgsin}\mathrm{\theta }\text{and}\mathrm{f}=\mathrm{\mu R}=\mathrm{\mu mgcos}\mathrm{\theta }$

$\therefore \mathrm{\mu mgcos}\mathrm{\theta }=\mathrm{mgsin}\mathrm{\theta }$

$\mathrm{\mu }=\tan \mathrm{\theta }$   … (1)

Case (ii). The block is projected upward with initial velocity u and hence it experiences downward acceleration a. In this case 

$\mathrm{mgsin}\mathrm{\theta }+\mathrm{\mu mgcos}\mathrm{\theta }=\mathrm{ma}$

$\text{or }\mathrm{mgsin}\mathrm{\theta }+\mathrm{mgtan}\mathrm{\theta cos}\mathrm{\theta }=\mathrm{ma}$

$\mathrm{or} \mathrm{mg}(\sin \mathrm{\theta }+\sin \mathrm{\theta })=\mathrm{ma}$

$\therefore \mathrm{a}=2\mathrm{gsin}\mathrm{\theta } ...\left(2\right)$

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

$\begin{array}{rl}{\mathrm{v}}^2-{\mathrm{u}}^2& =2\mathrm{as}\\ 0-{\mathrm{u}}^2& =2(-2\mathrm{gsin}\mathrm{\theta })\mathrm{x}\\ \mathrm{x}& =\frac{{\mathrm{u}}^2}{4\mathrm{gsin}\mathrm{\theta }}\end{array}$