A block of mass m is projected up an inclined plane of inclination $\mathrm{\theta }$ with an initial velocity u. If the coefficient of kinetic friction between the block and the plane is $\mathrm{\mu } = \tan \mathrm{\theta }$, the distance up to which the block will rise up the plane, before coming to rest, is given by

detailed solution

Correct option is C

Refer to Fig. Since the block is projected upwards, it rises after overcoming two forces: (i) the component mg sin θ of the weight mg and (ii) the force of friction F= mg sin θ, both acting downwards. Therefore, the total downward acceleration isa=−gsin⁡θ−gsin⁡θ=−2gsin⁡θLets be the distance moved up the plane before the block comes to rest. Then, from v2−u2=2as, we have0−u2=2×(−2gsin⁡θ)×s   or   s=u24gsin⁡θHence, the correct choice is (3).