A plane surface is inclined making an angle $\beta$ above the horizon. A bullet is fired with the point of projection at the bottom of the inclined plane with a velocity u; then the maximum range is given by:

detailed solution

Correct option is B

Suppose the particle be projected with a velocity u making an angle θ with the horizontal. Suppose the particle strikes the inclined plane at A after time T. If x-axis is taken along the inclined plane and y-axis normal to the inclined plane. Thenux = u cos (θ-β), uy = u sin(θ-β)During the time of flight T, the displacement along y-axis is zero. Using equation sy = uyt+12ayt20 = u sin(θ-β)T+12(-g cos β)T2T = 2u sin(θ-β)g cos βComponent of velocity along horizontal = u cos θDistance covered along horizontalOB = (u cos θ)T        = u cos θ[2u sin(θ-β)g cos β]        = 2u2cos θ sin(θ-β)g cos βRange R = OA OA = OBcos β =2u2 cos θ sin(θ-β)g cos2βR = 2u2 cos θ sin(θ-β)g cos2β   = [u2g cos2β][2 sin(θ-β)cos θ] = u2g cos2 β[sin(2θ-β)-sinβ]For a given u and β, R is maximum, when sin(2θ-β) is maximum, i.e., when   sin(2θ-β) = 1 or 2θ - β = π2or 2θ = β+π2 or θ = β2+π2Hence, the angle of projection θ, for maximum range up the inclined plane is given byθ = β2+π4Rmax = u2g cos2β[1-sin β]         = u2(1-sin β)g(1-sin2β) = u2g(1+sin β)