A projectile is thrown with a speed ‘u’ at angle 'θ' to an inclined plane of inclination β. The angle 'θ'  at which the projectile is thrown such that it strikes the inclined plane horizontally is

Taking horizontal as X-axis aand vertical as Y-axis, If particle strikes the plane horizontally then  its vertical velocity w.r.t ground is zero⇒vx=0 vx=ux+at 0=u sin(θ+β)-gt time=t=usinθ+βg time of flight on incine plane, T=2usinθgcosβ equating with time of flight of projection on inclined plane 2usinθgcosβ=usinθ+βg 2sinθ= sin(θ+β)cosβ let (θ+β)=k θ=k−β2sin⁡(k−β)=sin⁡k⋅cos⁡β2sin⁡k⋅cos⁡β−2cos⁡k⋅sin⁡β=sin⁡k⋅cos⁡βtan⁡k=2tan⁡βsubstitute k valuetan⁡(θ+β)=2tan⁡βtan⁡(θ+β)=2tan⁡β)θ+β=tan−1⁡(2tan⁡β)θ=tan−1⁡(2tan⁡β)−β