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

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).