A block slides down a slope of angle θ with constant velocity. It is then projected up with a velocity of 10 ms-1, g =10 ms-2 & θ = 30º. The maximum distance it can go up the plane before coming to stop is

Since the block slides down the inclined plane with constant velocity, mgsinθ = fk =µk.mgcosθ. During ascent, retardation (a) of the block is a = g(sinθ+μkcosθ)=2gsinθNow, 0 = 102 - 2 x 2gsinθ x s ⇒ S = 1004gsin300 = 5m