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JEE Main 26 Feb 2021 Shift 2 Solved Paper

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Question : 6

Total: 89

The trajectory of a projectile in a vertical plane is y=αx−βx2, where α and β are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by

[26 Feb 2021 Shift 2]

tan−1α,‌
α2
4β
tan−1β,‌
α2
2β
tan−1α,‌
4α2
β
tan−1(‌
β
α
),‌
α2
β

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