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Q.

A particle is projected with a certain velocity at an angle α above the horizontal from the foot of an inclined plane of inclination 30o. If the particle strikes the plane normally, then α is equal to

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a

30∘+tan−1⁡32

b

45∘

c

60∘

d

30∘+tan−1⁡(23)

answer is A.

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Detailed Solution

tAB= time of flight of projectile =2usin⁡α−30∘gcos⁡30∘Now the component of velocity along the plane becomes zero at point B.θ=ucos⁡α−30∘−gsin⁡30∘×tABor  ucos⁡α−30∘=gsin⁡30∘×2usin⁡α−30∘gcos⁡30∘or  tan⁡α−30∘=cot⁡30∘2=32α=30∘+tan−1⁡32
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