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

a
cot−1(tanβ)
b
cot−1(2tanβ)
c
sin−1(2tanβ)
d
sin−1(2cosβ)

detailed solution

Correct option is B

Time of flight=T=2usinθgcosβ                  _ _ _ _(1) ‘If it strikes the inclined plane normally  then component of velocity along the plane =vx= 0 we know,     vx=ux+at  subtitute vx=ucosθ−(gsinβ)t t=ucosθgsinβ                                            −−−−−− 2  Equating (1) & (2)  2usinθgcosβ=ucosθgsinβ 2tanβ = cotθ θ= cot−1(2tanβ)

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