A projectile is thrown with a speed ‘u’ at angle $\text{'}\theta \text{'}$ to an inclined plane of inclination $\beta$. The angle $\text{'}\theta \text{'}$  at which the projectile is thrown such that it strikes the inclined plane horizontally is

detailed solution

Correct option is A

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