The trajectory of a projectile in a vertical plane is y=αx−βx2, where α and β are constants and x & 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 :
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a
tan−1α,4α2β
b
tan−1βα,α2β
c
tan−1α,α24β
d
tan−1β,α22β
answer is C.
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Detailed Solution
y=αx−βx2 Compare with standard form, y=tanθx−g2u2cos2θx2 We get, tanθ=α ⇒θ=tan−1α and α24β=tan2θ4g2u2cos2θ=H