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Q.

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
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